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CBSE Class 9 Mathematics Question 5 of 16

Polynomials — Question 4

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4
Question

Question 4

Expand 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

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

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

Putting x = a4,y=−b2\dfrac{a}{4}, y = -\dfrac{b}{2}, 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}

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Polynomials — Interactive Study Guide

Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification

Not a polynomial: √x (fractional power), 1/x = x−1 (negative power), x + 1/x.

Polynomial: 5 (constant), 3x + 2 (linear), x² − 1 (quadratic), 2x³ + x − 1 (cubic).

Remainder and Factor Theorems — Quick Guide

Remainder Theorem: When p(x) is divided by (x − a), remainder = p(a).
Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.

Watch the sign! Dividing by (x + 3) means a = −3. So remainder = p(−3).

Identity Mastery Checklist

See This PatternUse This Identity
a² + 2ab + b²= (a + b)²
a² − 2ab + b²= (a − b)²
a² − b²= (a + b)(a − b)
a³ + b³= (a + b)(a² − ab + b²)
a³ − b³= (a − b)(a² + ab + b²)
a + b + c = 0⇒ a³ + b³ + c³ = 3abc

Quick Self-Check

  1. Degree of 5x³ − 2x + 1? (3)
  2. Remainder when x² + 3x + 2 is divided by (x + 1)? (p(−1) = 1 − 3 + 2 = 0)
  3. Expand: (2a + 3b)² (= 4a² + 12ab + 9b²)
  4. Factorise: 8x³ − 27 (= (2x − 3)(4x² + 6x + 9))

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