Question 1

Question 2Write the coefficients of x2 in each of the following:(i) 2 + x2 + x(ii) 2 - x2 + x3(iii) π2\dfrac{π}{2}2π​ x2 + x(iv) 2\sqrt{2}2​x - 1

Accepted Answer

(i) 2 + x2 + xCoefficient of x2 is 1.(ii) 2 - x2 + x3Coefficient of x2 is (-1).(iii) π2\dfrac{π}{2}2π​ x2 + xCoefficient of x2 is = π2\dfrac{π}{2}2π​(iv) 2\sqrt{2}2​x - 1Coefficient of x2 is 0.

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 2

Question 3Verify whether the following are zeroes of the polynomial, indicated against them.(i) p(x) = 3x + 1, x = −13-\dfrac{1}{3}−31​(ii) p(x) = 5x - π, x = 45\dfrac{4}{5}54​(iii) p(x) = x2 - 1, x = 1, -1(iv) p(x) = (x + 1)(x - 2), x = -1, 2(v) p(x) = x2, x = 0(vi) p(x) = lx + m, x = −ml-\dfrac{m}{l}−lm​(vii) p(x) = 3x2 - 1, x = −(13)-\Big(\dfrac{1}{\sqrt{3}}\Big)−(3​1​), (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​)(viii) p(x) = 2x + 1, x = 12\dfrac{1}{2}21​

Accepted Answer

(i) p(x) = 3x + 1Putting x = −13-\dfrac{1}{3}−31​ we get,p(−13)	ext{p}\Big(-\dfrac{1}{3}\Big)p(−31​) = 3(−13)3\Big(-\dfrac{1}{3}\Big)3(−31​) + 1= -1 + 1= 0Hence, x = −13-\dfrac{1}{3}−31​ is zeroes of this polynomial.(ii) p(x) = 5x - πPutting x = 45\dfrac{4}{5}54​ we get,p(45)=5(45)−π	ext{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pip(54​)=5(54​)−π= 4 - πHence, x = 45\dfrac{4}{5}54​ is not a zeroes of this polynomial.(iii) p(x) = x2 - 1Putting x = 1 we get,p(1) = (1)2 - 1= 0Putting x = -1 we getp(-1) = (-1)2 - 1= 1 - 1= 0Hence, x = 1, -1 is zeroes of this polynomial.(iv) p(x) = (x + 1)(x - 2)Putting x = -1 we get,p(1) = (-1 + 1)(-1 - 2)= 0 x (-3)= 0Putting x = 2 we get,p(2) = (2 + 1)(2 - 2)= 3 x 0= 0Hence, x = -1, 2 are the zeroes of this polynomial.(v) p(x) = x2Putting x = 0 we get,p(0) = (0)2= 0Hence, x = 0 is the zero of this polynomial.(vi) p(x) = lx + mPutting x = -ml\dfrac{m}{l}lm​ we get,p - (ml)=lx−(ml)\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big)(lm​)=lx−(lm​) + m= -m + m= 0Hence, x = −ml-\dfrac{m}{l}−lm​ is the zero of this polynomial.(vii) p(x) = 3x2 - 1Putting x = −13-\dfrac{1}{\sqrt{3}}−3​1​ we get,p-(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(3​1​) = 3 x -(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(3​1​)2 - 1= 3 x 13\dfrac{1}{3}31​ - 1= 1 - 1= 0Putting x = 23\dfrac{2}{\sqrt{3}}3​2​ we get,p (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​) = 3 x (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​)2 - 1= 3 x 43\dfrac{4}{3}34​ - 1= 4 - 1= 3Hence, x = −13-\dfrac{1}{\sqrt{3}}−3​1​ is zeroes of polynomial but x = 23\dfrac{2}{\sqrt{3}}3​2​ is not a zeroes of this polynomial.(viii) p(x) = 2x + 1Putting x = 12\dfrac{1}{2}21​ we get,p(12)\Big(\dfrac{1}{2}\Big)(21​) = 2 x 12\dfrac{1}{2}21​ + 1= 1 + 1= 2No, x = 12\dfrac{1}{2}21​ is not the zeroes of this polynomial.

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 3

Question 1Determine which of the following polynomials has (x + 1) a factor :(i) x3 + x2 + x + 1(ii) x4 + x3 + x2 + x + 1(iii) x4 + 3x3 + 3x2 + x + 1(iv) x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​

Accepted Answer

(i) x3 + x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x3 + x2 + x + 1p(-1) = (-1)3 + (-1)2 + (-1) + 1= -1 + 1 -1 + 1= 0Remainder is zero (0), so (x + 1) is factor of this polynomial.(ii) x4 + x3 + x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x4 + x3 + x2 + x + 1p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) + 1= 1 -1 + 1 -1 + 1= 1Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.(iii) x4 + 3x3 + 3x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x4 + 3x3 + 3x2 + x + 1p(-1) = (-1)4 + 3 x (-1)3 + 3 x (-12) + (-1) + 1= 1 -3 + 3 -1 + 1= 1Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.(iv) x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​⇒ x + 1 = 0⇒ x = -1p(x) = x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​p(-1) = (-1)3 - (-1)2 - (2 + 2\sqrt{2}2​)(-1) + 2\sqrt{2}2​= -1 - 1 + 2 + 2\sqrt{2}2​ + 2\sqrt{2}2​= -2 + 2 + 2 2\sqrt{2}2​= 2 2\sqrt{2}2​Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 4

Question 5Factorise :(i) x3 - 2x2 - x + 2(ii) x3 - 3x2 - 9x - 5(iii) x3 + 13x2 + 32x + 20(iv) 2y3 + y2 - 2y - 1

Accepted Answer

(i) x3 - 2x2 - x + 2Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = 1x - 1 = 0x = 1p(x) = x3 - 2x2 - x + 2p(1) = (1)3 - 2 x (1)2 - 1 + 2= 1 - 2 - 1 + 2= 0Hence, (x - 1) is the factor of x3 - 2x2 - x + 2On dividing, x3 - 2x2 - x + 2 by (x - 1)x−1)x2−x−2x−1)x3−2x2−x+2‾x−1)2+−x3−+x2‾x−1x3−2−x2−x+2x−1)x3−2−+x2+−x‾x−1)x3−2x2(3)−2x+2x−1)x3−2x2(31)−+2x+−2‾x−1)x3−2x2(31)−2x×\begin{array}{l} \phantom{x - 1)}{\quad x^2 -x - 2} \ x - 1\overline{\smash{\big)}\quad x^3 - 2x^2 - x + 2} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{+}{-}x^2} \ \phantom{{x - 1}x^3-2}-x^2 - x + 2 \ \phantom{{x - 1)}x^3-2}\underline{\underset{+}{-}x^2 \underset{-}{+} x} \ \phantom{{x - 1)}{x^3-2x^{2}(3)}}-2x + 2 \ \phantom{{x - 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 2} \ \phantom{{x - 1)}{x^3-2x^{2}(31)}{-2x}}	imes \end{array}x−1)x2−x−2x−1)x3−2x2−x+2​x−1)2−+​x3+−​x2​x−1x3−2−x2−x+2x−1)x3−2+−​x2−+​x​x−1)x3−2x2(3)−2x+2x−1)x3−2x2(31)+−​2x−+​2​x−1)x3−2x2(31)−2x×​we get quotient = x2 - x - 2Factorising x2 - x - 2,= x2 - x - 2= x2 -(2 - 1)x - 2= x2 - 2x + x -2= x( x - 2) + 1(x + 2)= (x + 1)(x - 2)∴ x2 - x - 2 = (x + 1)(x - 2)∴ p(x) = x3 - 2x2 - x + 2 = (x - 1) (x + 1) (x - 2)So, factors are (x - 1) (x + 1) (x - 2)(ii) x3 - 3x2 - 9x - 5Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = -1x + 1 = 0x = -1p(x) = x3 - 3x2 - 9x - 5p(-1) = (-1)3 - 3 x (-1)2 - 9 x (-1) - 5= -1 - 3 + 9 - 5= -9 + 9= 0Hence, (x + 1) is a factor of x3 - 3x2 - 9x - 5On dividing, x3 - 3x2 - 9x - 5 by (x + 1)x+1)x2−4x−5x+1)x3−3x2−9x−5‾x−1)2+−x3+−x2‾x+1x3−2−4x2−9xx+1)x3−2−+4x2−+4x‾x+1)x3−2x2(3)−5x−5x+1)x3−2x2(31)−+5x−+5‾x+1)x3−2x2(31)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 -4x - 5} \ x + 1\overline{\smash{\big)}\quad x^3 - 3x^2 - 9x - 5} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}-4x^2 - 9x \ \phantom{{x + 1)}x^3-2}\underline{\underset{+}{-}4x^2 \underset{+}{-} 4x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}-5x - 5 \ \phantom{{x + 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}5x \underset{+}{-} 5} \ \phantom{{x + 1)}{x^3-2x^{2}(31)}{-2}}	imes \end{array}x+1)x2−4x−5x+1)x3−3x2−9x−5​x−1)2−+​x3−+​x2​x+1x3−2−4x2−9xx+1)x3−2+−​4x2+−​4x​x+1)x3−2x2(3)−5x−5x+1)x3−2x2(31)+−​5x+−​5​x+1)x3−2x2(31)−2×​We get quotient = x2 - 4x - 5Factorising x2 - 4x - 5x2 -(5 - 1)x - 5x2 -5x + x - 5x(x - 5) + 1(x - 5)(x + 1) (x - 5)∴ x2 - 4x - 5 = (x + 1) (x - 5)∴ p(x) = x3 - 3x2 -9x - 5 = (x + 1) (x + 1) (x - 5)So, factors are (x + 1) (x + 1) (x - 5)(iii) x3 + 13x2 + 32x + 20Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = -1x + 1 = 0x = -1p(x) = x3 + 13x2 + 32x + 20p(-1) = (-1)3 + 13 x (-1)2 + 32 x (-1) + 20= -1 + 13 - 32 + 20= -33 + 33= 0Hence, (x + 1) is a factor of x3 + 13x2 + 32x + 20On dividing, x3 + 13x2 + 32x + 20 by (x + 1)x+1)x2+12x+20x+1)x3+13x2+32x+20‾x−1)2+−x3+−x2‾x+1x3−212x2+32x+20x+1)x31+−12x2+−12x‾x+1)x3−2x2(3)20x+20x+1)x3−2x23+−20x+−20‾x+1)x3−2x2(3)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 +12x + 20} \ x + 1\overline{\smash{\big)}\quad x^3 + 13x^2 + 32x + 20} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}12x^2 + 32x + 20 \ \phantom{{x + 1)}x^31}\underline{\underset{-}{+}12x^2 \underset{-}{+} 12x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}20x + 20 \ \phantom{{x + 1)}{x^3-2x^{2}3}}\underline{\underset{-}{+}20x \underset{-}{+} 20} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}{-2}}	imes \end{array}x+1)x2+12x+20x+1)x3+13x2+32x+20​x−1)2−+​x3−+​x2​x+1x3−212x2+32x+20x+1)x31−+​12x2−+​12x​x+1)x3−2x2(3)20x+20x+1)x3−2x23−+​20x−+​20​x+1)x3−2x2(3)−2×​We get quotient = x2 + 12x + 20Factorising x2 + 12x + 20x2 + 10x + 2x + 20x(x + 10) + 2(x + 10)(x + 10)(x + 2)∴ x2 + 12x + 20 = (x + 10)(x + 2)∴ p(x) = x3 + 13x2 + 32x + 20 = (x + 1) (x + 10) (x + 2)So, factors are (x + 1) (x + 10) (x + 2)(iv) 2y3 + y2 - 2y - 1Let (y - a) is a factor, a = 1, -1, 2, -2.......Putting a = 1y - 1 = 0y = 1p(y) = 2y3 + y2 - 2y - 1p(1) = 2 x (1)3 + (1)2 - 2 x 1 - 1= 2 + 1 - 2 - 1= 0Hence, (y - 1) is a factor of 2y3 + y2 - 2y - 1On dividing, 2y3 + y2 - 2y - 1 by (y - 1)y−1)2y2+3y+1y−1)2y3+y2−2y−1‾y−1)+−2y3−+2y2‾y−1x3−2333y2−2yy−1)x3−2+−3y2−+3y‾y−1)x3−2x2(3333)y−1y−1)x3−2x23333+−y−+1‾y−1)x3−2x2(333)−2×\begin{array}{l} \phantom{y - 1)}{\quad 2y^2 + 3y + 1} \ y - 1\overline{\smash{\big)}\quad 2y^3 + y^2 - 2y - 1} \ \phantom{y - 1)}\underline{\underset{-}{+}2y^3 \underset{+}{-}2y^2} \ \phantom{{y - 1}x^3-233}3y^2 - 2y \ \phantom{{y - 1)}x^3-2}\underline{\underset{-}{+}3y^2 \underset{+}{-} 3y} \ \phantom{{y - 1)}{x^3-2x^{2}(3333)}}y - 1 \ \phantom{{y - 1)}{x^3-2x^{2}3333}}\underline{\underset{-}{+}y \underset{+}{-} 1} \ \phantom{{y - 1)}{x^3-2x^{2}(333)}{-2}}	imes \end{array}y−1)2y2+3y+1y−1)2y3+y2−2y−1​y−1)−+​2y3+−​2y2​y−1x3−2333y2−2yy−1)x3−2−+​3y2+−​3y​y−1)x3−2x2(3333)y−1y−1)x3−2x23333−+​y+−​1​y−1)x3−2x2(333)−2×​We get quotient = 2y2 + 3y + 1Factorising 2y2 + 3y + 12y2 + 2y + y +12y(y + 1) + 1(y + 1)(y + 1)(2y + 1)∴ 2y2 + 3y + 1 = (y + 1)(2y + 1)∴ p(y) = 2y3 + y2 - 2y - 1 = (y - 1)(y + 1)(2y + 1)So, factors are (y - 1)(y + 1)(2y + 1)

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 5

Question 4Expand each of the following, using suitable identities:(i) (x + 2y + 4z)2(ii) (2x - y + z)2(iii) (-2x + 3y + 2z)2(iv) (3a - 7b - c)2(v) (-2x + 5y - 3z)2(vi) [14a−12b+1]2\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2[41​a−21​b+1]2

Accepted Answer

(i) (x + 2y + 4z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = x, y = 2y, z = 4z= (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx(ii) (2x - y + z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = 2x, y = (-y), z = z= (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)= 4x2 + y2 + z2 - 4xy - 2yz + 4zx(iii) (-2x + 3y + 2z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = -2x, y = 3y, z = 2z= (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8zx(iv) (3a - 7b - c)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = 3a, y = (-7b), z = (-c)= (3a)2 + (-7b)2 + (-c)2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)= 9a2 + 49b2 + c2 - 42ab + 14bc - 6ac(v) (-2x + 5y - 3z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = -2x, y = 5y, z = (-3z)= (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12zx(vi) [14a−12b+1]2\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2[41​a−21​b+1]2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = a4,y=−b2\dfrac{a}{4}, y = -\dfrac{b}{2}4a​,y=−2b​, z = 1=(a4)2+(−b2)2+(1)2+2(a4)(−b2)+2(−b2)(1)+2(1)(a4)=a216+b24+1−ab4−b+a2= \Big(\dfrac{a}{4}\Big)^2 + \Big(-\dfrac{b}{2}\Big)^2 + (1)^2 + 2\Big(\dfrac{a}{4}\Big) \Big(-\dfrac{b}{2}\Big) + 2\Big(-\dfrac{b}{2}\Big)(1) + 2(1) \Big(\dfrac{a}{4}\Big) \[1em] = \dfrac{a^2}{16} + \dfrac{b^2}{4} + 1 - \dfrac{ab}{4} -b + \dfrac{a}{2}=(4a​)2+(−2b​)2+(1)2+2(4a​)(−2b​)+2(−2b​)(1)+2(1)(4a​)=16a2​+4b2​+1−4ab​−b+2a​

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 6

Question 6Write the following cubes in expanded form:(i) (2x + 1)3(ii) (2a - 3b)3(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3[23x​+1]3(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3[x−32​y]3

Accepted Answer

(i) (2x + 1)3We know that:(a + b)3= (a)3 + (b)3 + 3a2b + 3ab2Putting a = 2x, b = 1= (2x)3 + (1)3 + 3(2x)2(1) + 3(2x)(1)2= 8x3 + 12x2 + 6x + 1Hence, (2x + 1)3 = 8x3 + 12x2 + 6x + 1(ii) (2a - 3b)3We know that:(a - b)3= (a)3 - (b)3 - 3a2b + 3ab2Putting a = 2a, b = 3b= (2a)3 - (3b)3 - 3(2a)2(3b) + 3(2a)(3b)2= 8a3 - 27b3 - 3(4a2)(3b) + 3(2a)(9b2)= 8a3 - 27b3 - 36a2b + 54ab2Hence, (2a - 3b)3 = 8a3 - 27b3 - 36a2b + 54ab2(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3[23x​+1]3We know that:(a + b)3 = (a)3 + (b)3 + 3ab(a + b)Putting a = 32x\dfrac{3}{2}x23​x, b = 1=(32)3+(1)3+3(3x2)(1)(3x2+1)=27x38+1+(9x2)(3x2+1)=27x38+1+(9x2)(3x2)+9x2(1)=27x38+27x24+9x2+1= \Big(\dfrac{3}{2}\Big)^3 + (1)^3 + 3 \Big(\dfrac{3x}{2}\Big)(1) \Big(\dfrac{3x}{2} + 1\Big) \[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2} + 1\Big) \[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2}\Big) + \dfrac{9x}{2}(1)\[1em] = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1=(23​)3+(1)3+3(23x​)(1)(23x​+1)=827x3​+1+(29x​)(23x​+1)=827x3​+1+(29x​)(23x​)+29x​(1)=827x3​+427x2​+29x​+1Hence, [3x2+1]3=27x38+27x24+9x2+1\Big[\dfrac{3x}{2} + 1\Big]^3 = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1[23x​+1]3=827x3​+427x2​+29x​+1(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3[x−32​y]3We know that:(a - b)3 = (a)3 - (b)3 - 3ab(a - b)Putting a = x, b = −23y-\dfrac{2}{3}y−32​y=(x)3−[−23y]3−3(x)[−23y][x−23y]=x3−827y3−2xy[x−23y]=x3−827y3−2x2y+43xy2= (x)^3 - \Big[-\dfrac{2}{3}y\Big]^3 - 3(x)\Big[-\dfrac{2}{3}y\Big] \Big[x - \dfrac{2}{3}y\Big]\[1em] = x^3 - \dfrac{8}{27}y^3 - 2xy\Big[x - \dfrac{2}{3}y\Big]\[1em] = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2=(x)3−[−32​y]3−3(x)[−32​y][x−32​y]=x3−278​y3−2xy[x−32​y]=x3−278​y3−2x2y+34​xy2Hence, [x−23y]3=x3−827y3−2x2y+43xy2\Big[x - \dfrac{2}{3}y\Big]^3 = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2[x−32​y]3=x3−278​y3−2x2y+34​xy2

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 7

Question 7Evaluate the following using suitable identities:(i) (99)3(ii) (102)3(iii) (998)3

Accepted Answer

(i) (99)3We can write it as,= (100 - 1)3We know that,(a - b)3 = (a)3 - (b)3 - 3ab(a - b)Putting a = 100 , b = 1= (100)3 - (1)3 - 3 x 100 x 1(100 - 1)= 1000000 - 1 - 300 x 99= 1000000 - 1 - 29700= 970299Hence, (99)3 = 970299(ii) (102)3We can write it as,= (100 + 2)3We know that,(a + b)3 = (a)3 + (b)3 + 3ab(a + b)Putting a = 100 , b = 2= (100)3 + (2)3 + 3 x 100 x 2(100 + 2)= 1000000 + 8 + 600 x 102= 1000000 + 8 + 61200= 1061208Hence, (102)3 = 1061208(iii) (998)3We can write it as,= (1000 - 2)3We know that,(a - b)3 = (a)3 - (b)3 - 3ab(a - b)Putting a = 1000 , b = 2= (1000)3 - (2)3 - 3 x 1000 x 2(1000 - 2)= 1000000000 - 8 - 6000 x 998= 1000000000 - 8 - 5988000= 994011992Hence, (998)3 = 994011992

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 8

Question 8Factorise each of the following:(i) 8a3 + b3 + 12a2b + 6ab2(ii) 8a3 - b3 - 12a2b + 6ab2(iii) 27 - 125a3 - 135a + 225a2(iv) 64a3 - 27b3 - 144a2b + 108ab2(v) 27p3 - 1216\dfrac{1}{216}2161​ - 92\dfrac{9}{2}29​p2 + 14\dfrac{1}{4}41​p

Accepted Answer

(i) 8a3 + b3 + 12a2b + 6ab2[∵ a3 + b3 + 3a2b + 3ab2 = (a + b)3]= (2a)3 + (b)3 + 3(2a)2(b) + 3(2a)(b)2= (2a + b)3Hence, 8a3 + b3 + 12a2b + 6ab2 = (2a + b)(2a + b)(2a + b)(ii) 8a3 - b3 - 12a2b + 6ab2[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2= (2a - b)3Hence, 8a3 - b3 - 12a2b + 6ab2 = (2a - b)(2a - b)(2a - b)(iii) 27 - 125a3 - 135a + 225a2[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]= (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2= (3 - 5a)3Hence, 27 - 125a3 - 135a + 225a2 = (3 - 5a)(3 - 5a)(3 - 5a)(iv) 64a3 - 27b3 - 144a2b + 108ab2[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2= (4a - 3b)3Hence, 64a3 - 27b3 - 144a2b + 108ab2 = (4a - 3b)(4a - 3b)(4a - 3b)(v) 27p3 - 1216\dfrac{1}{216}2161​ - 92\dfrac{9}{2}29​p2 + 14\dfrac{1}{4}41​p[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]= (3p)3 - (16)\Big(\dfrac{1}{6}\Big)(61​)3 - 3(3p)2(16)\Big(\dfrac{1}{6}\Big)(61​) + 3(3p)(16)\Big(\dfrac{1}{6}\Big)(61​)2= (3p−16)3\Big(3p - \dfrac{1}{6}\Big)^3(3p−61​)3Hence, 27p3 - 1216\dfrac{1}{216}2161​ - 92\dfrac{9}{2}29​p2 + 14\dfrac{1}{4}41​p = (3p−16)\Big(3p - \dfrac{1}{6}\Big)(3p−61​)\Big(3p - \dfrac{1}{6}\Big)(3p−16)\Big(3p - \dfrac{1}{6}\Big)(3p−61​)

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 9

Question 9Verify :(i) x3 + y3 = (x + y) (x2 - xy + y2)(ii) x3 - y3 = (x - y) (x2 + xy + y2)

Accepted Answer

(i) x3 + y3 = (x + y) (x2 - xy + y2)R.H.S= (x + y) (x2 - xy + y2)= (x)(x2 - xy + y2) + (y)(x2 - xy + y2)= x3 - x2y + xy2 + yx2 - xy2 + y3= x3 + y3Hence, L.H.S = R.H.S(ii) x3 - y3 = (x - y) (x2 + xy + y2)R.H.S= (x - y) (x2 + xy + y2)= (x)(x2 + xy + y2) - (y)(x2 + xy + y2)= x3 + x2y + xy2 - yx2 - xy2 - y3= x3 - y3Hence, L.H.S = R.H.S

Polynomials - Interactive Study Notes | Bright Tutorials

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    .quick-quiz h4 { color: #e67e22; margin-top: 0; }

BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 10

Question 10Factorise each of the following:(i) 27y3 + 125z3(ii) 64m3 - 343n3

Accepted Answer

(i) 27y3 + 125z3[∵ x3 + y3 = (x + y)(x2 - xy + y2)]⇒ 27y3 + 125z3 = (3y)3 + (5z)3Putting x = 3y, y = 5zR.H.S = (x + y)(x2 - xy + y2)= (3y + 5z)[(3y)2 - (3y)(5z) + (5z)2]= (3y + 5z)(9y2 - 15yz + 25z2)Hence, 27y3 + 125z3 = (3y + 5z)(9y2 - 15yz + 25z2)(ii) 64m3 - 343n3[∵ a3 - b3 = (a - b)(a2 + ab + b2)]⇒ 64m3 - 343n3 = (4m)3 - (7n)3Putting a = 4m, b = 7nR.H.S = (a - b)(a2 + ab + b2)= (4m - 7n)[(4m)2 + (4m)(7n) + (7n)2]= (4m - 7n)(16m2 + 28mn + 49n2)Hence, 64m3 - 343n3 = (4m - 7n)(16m2 + 28mn + 49n2)

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 11

Question 11Factorise : 27x3 + y3 + z3 - 9xyz

Accepted Answer

27x3 + y3 + z3 - 9xyz[∵ x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)]= (3x)3 + y3 + z3 - 3(3x)(y)(z)= (3x + y + z)[(3x)2 + y2 + z2 - (3x)(y) - (y)(z) - (z)(3x)]= (3x + y + z)(9x2 + y2 + z2 - 3xy - yz - 3zx)Hence, 27x3 + y3 + z3 - 9xyz = (3x + y + z)(9x2 + y2 + z2 - 3xy - yz - 3zx)

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 12

Question 12Verify that x3 + y3 +z3 - 3xyz = 12\dfrac{1}{2}21​(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2]

Accepted Answer

R.H.S= 12\dfrac{1}{2}21​(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2]= 12\dfrac{1}{2}21​(x + y + z)[(x2 + y2 - 2xy) + (y2 + z2 - 2yz) + (z2 + x2 - 2zx)] [∵ (a - b)2 = (a)2 + (b)2 - 2ab]= 12\dfrac{1}{2}21​(x + y + z)[2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx]= 12\dfrac{1}{2}21​(x + y + z) 2[x2 + y2 + z2 - xy - yz - zx]= (x + y + z)[x2 + y2 + z2 - xy - yz - zx]= x3 + y3 +z3 - 3xyz [∵ x3 + y3 +z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)]Hence, L.H.S = R.H.S

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 13

Question 13If x + y + z = 0, show that x3 + y3 + z3 = 3xyz

Accepted Answer

(x + y + z)3 = x3 + y3 + z3 - 3xyzGiven that x + y + z = 0= (0)3 = x3 + y3 + z3 - 3xyz⇒ 3xyz = x3 + y3 + z3Hence proved

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 14

Question 14Without actually calculating the cubes, find the value of each of the following:(i) (-12)3 + (7)3 + (5)3(ii) (28)3 + (-15)3 + (-13)3

Accepted Answer

(i) (-12)3 + (7)3 + (5)3Here,-12 + 7 + 5 = 0As, x + y + z = 0So, x3 + y3 + z3 = 3xyz= 3 x (-12) x 7 x 5= -1260Hence, (-12)3 + (7)3 + (5)3 = -1260(ii) (28)3 + (-15)3 + (-13)3Here,= 28 + (-15) + (-13)= 28 - 15 - 13= 0As, x + y + z = 0So, x3 + y3 + z3 = 3xyz= 3 x 28 x (-15) x (-13)= 16380Hence, (28)3 + (-15)3 + (-13)3 = 16380

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 15

Question 15Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:(i) Area: 25a2 - 35a + 12(ii) Area: 35y2 + 13y - 12

Accepted Answer

(i) 25a2 - 35a + 12Area of Rectangle = Length x Breadth= 25a2 - 35a + 12= 25a2 -(20a + 15a) + 12= 25a2 -20a -15a + 12= 5a(5a - 4) - 3(5a - 4)= (5a - 4)(5a - 3)Hence, one possible answer is : Length = 5a - 3, Breadth = 5a - 4(ii) 35y2 + 13y - 12Area of Rectangle = Length x Breadth= 35y2 + 13y - 12= 35y2 +(28y - 15y) - 12= 35y2 + 28y - 15y - 12= 7y(5y + 4) - 3(5y + 4)= (5y + 4)(7y - 3)Hence, one possible answer is : Length = 7y - 3, Breadth = 5y + 4

Polynomials - Interactive Study Notes | Bright Tutorials

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    .quick-quiz h4 { color: #e67e22; margin-top: 0; }

BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 16

Question 16What are the possible expressions for the dimensions of the cuboids whose volume are given below?(i) Volume: 3x2 - 12x(ii) Volume: 12ky2 + 8ky - 20k

Accepted Answer

(i) Volume: 3x2 - 12xVolume of Cuboid = Length x Breadth X Height= 3x2 - 12x= 3x(x - 4)= 3(x)(x - 4)Hence, one possible answer is : 3, x and x - 4.(ii) Volume: 12ky2 + 8ky - 20k= 12ky2 + 8ky - 20k= 4k(3y2 + 2y - 5)= 4k(3y2 + 5y - 3y - 5)= 4k[y(3y + 5) -1(3y + 5)]= 4k (y - 1)(3y + 5)Hence, one possible answer is : 4k, 3y + 5 and y - 1.

Polynomials - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | PolynomialsWeb Content &bull; Interactive Notes
  
    Polynomials &mdash; Interactive Study Guide
    Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification
      Not a polynomial: &radic;x (fractional power), 1/x = x&minus;1 (negative power), x + 1/x.
      Polynomial: 5 (constant), 3x + 2 (linear), x&sup2; &minus; 1 (quadratic), 2x&sup3; + x &minus; 1 (cubic).

Remainder and Factor Theorems &mdash; Quick Guide
      Remainder Theorem: When p(x) is divided by (x &minus; a), remainder = p(a).
      Factor Theorem: (x &minus; a) is a factor of p(x) &hArr; p(a) = 0.
      Watch the sign! Dividing by (x + 3) means a = &minus;3. So remainder = p(&minus;3).

Identity Mastery Checklist
      
        See This PatternUse This Identity
        
          a&sup2; + 2ab + b&sup2;= (a + b)&sup2;
          a&sup2; &minus; 2ab + b&sup2;= (a &minus; b)&sup2;
          a&sup2; &minus; b&sup2;= (a + b)(a &minus; b)
          a&sup3; + b&sup3;= (a + b)(a&sup2; &minus; ab + b&sup2;)
          a&sup3; &minus; b&sup3;= (a &minus; b)(a&sup2; + ab + b&sup2;)
          a + b + c = 0&rArr; a&sup3; + b&sup3; + c&sup3; = 3abc

Quick Self-Check
      
        Degree of 5x&sup3; &minus; 2x + 1? (3)
        Remainder when x&sup2; + 3x + 2 is divided by (x + 1)? (p(&minus;1) = 1 &minus; 3 + 2 = 0)
        Expand: (2a + 3b)&sup2; (= 4a&sup2; + 12ab + 9b&sup2;)
        Factorise: 8x&sup3; &minus; 27 (= (2x &minus; 3)(4x&sup2; + 6x + 9))

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Polynomials

Exercise 21

Exercise 22

Exercise 23

Exercise 24