Factorise the following using appropriate identities: (i) 9x^2 + 6xy + y^2 (ii) 4y^2 - 4y + 1 (iii)x^2-(y^2/100)

(i) 9x 2 + 6xy + y 2 [∵ a 2 + 2ab + b 2 = (a + b) 2 ] = (3x) 2 + 2(3x)(y) + (y) 2 = (3x + y) 2 Hence, 9x 2 + 6xy + y 2 = (3x + y)(3x + y) (ii) 4y 2 - 4y + 1 [∵ a 2 - 2ab + b 2 = (a - b) 2 ] = (2y) 2 - 2(2y)(1) + (1) 2 = (2y - 1) 2 Hence, 4y 2 - 4y + 1 = (2y - 1)(2y - 1) (iii) [∵ a 2 - b 2 = (a - b)(a + b)] Hence, =

Mathematics

Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²
(ii) 4y² - 4y + 1
(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

Polynomials

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Answer

(i) 9x² + 6xy + y²

[∵ a² + 2ab + b² = (a + b)²]

= (3x)² + 2(3x)(y) + (y)²

= (3x + y)²

Hence, 9x² + 6xy + y² = (3x + y)(3x + y)

(ii) 4y² - 4y + 1

[∵ a² - 2ab + b² = (a - b)²]

= (2y)² - 2(2y)(1) + (1)²

= (2y - 1)²

Hence, 4y² - 4y + 1 = (2y - 1)(2y - 1)

(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

[∵ a² - b² = (a - b)(a + b)]

$x^2 - \Big(\dfrac{y^2}{100}\Big) \\[1em] = x^2 - \Big(\dfrac{y}{10}\Big)^2 \\[1em] = \Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)$

Hence, $x^2 - \Big(\dfrac{y^2}{100}\Big)$ = $\Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)$

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Mathematics

Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²
(ii) 4y² - 4y + 1
(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

Polynomials

Answer

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Mathematics

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Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²
(ii) 4y² - 4y + 1
(iii) $x^2 - \Big(\dfrac{y^2}{100}\Big)$

Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
(ii) (x + 8)(x - 10)
(iii) (3x + 4)(3x - 5)
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Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96

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) $\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2$

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(ii) 2x² + y² + 8z² - $2\sqrt{2}$ xy + $4\sqrt{2}$ yz - 8xz