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

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(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 +

Polynomials — Question 5

Question 5