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Factorisation Using Algebraic Identities

Use a suitabl...

Question

Use a suitable identity to get each of the following products.
(i) $(x + 3) (x + 3)$
(ii) $(2 y + 5) (2 y + 5)$
(iii) $(2 a - 7) (2 a - 7)$
(iv) $(\begin{matrix} 3 a - \frac{1}{2} \end{matrix}) (\begin{matrix} 3 a - \frac{1}{2} \end{matrix})$
(v) $(1.1 m - 0.4) (1.1 m + 0.4)$
(vi) $(a^{2} + b^{2}) (- a^{2} + b^{2})$
(vii) $(6 x - 7) (6 x + 7)$
(viii) $(- a + c) (- a + c)$
(ix) $(\begin{matrix} \frac{x}{2} + \frac{3 y}{4} \end{matrix}) (\begin{matrix} \frac{x}{2} + \frac{3 y}{4} \end{matrix})$
(x) $(7 a - 9 b) (7 a - 9 b)$

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Solution

(i) Given $(x + 3) (x + 3) = (x + 3)^{2}$ $∵ [a \times a = a^{2}]$
$= x^{2} + 3^{2} + 2 (x) (3)$ $∵ [(a + b)^{2} = a^{2} + b^{2} + 2 a b]$
$∴ (x + 3) (x + 3) = x^{2} + 9 + 6 x$

(ii) Given $(2 y + 5) (2 y + 5) = (2 y + 5)^{2}$ $∵ [a \times a = a^{2}]$
$= (2 y)^{2} + (5)^{2} + 2 (2 y) (5)$ $∵ [(a + b)^{2}$ $= a^{2} + b^{2} + 2 a b]$
$∴ (2 y + 5) (2 y + 5) = 4 y^{2} + 25 + 20 y$

(iii) Given $(2 a - 7) (2 a - 7) = (2 a - 7)^{2}$
$= (2 a)^{2} + (7)^{2} - 2 (2 a) (7)$ $∵ [(a - b)^{2}$ $= a^{2} + b^{2} - 2 a b]$
$∴ (2 a - 7) (2 a - 7) = 4 a^{2} + 49 - 28 a$
(iv) Given $(3 a - \frac{1}{2}) (3 a - \frac{1}{2}) = (3 a - \frac{1}{2})^{2}$
$= (3 a)^{2} + (\frac{1}{2})^{2} - (2) 3 a (\frac{1}{2})$ $∵ [(a - b)^{2}$ $= a^{2} + b^{2} - 2 a b]$
$∴ (3 a - \frac{1}{2}) (3 a - \frac{1}{2}) = 9 a^{2} + \frac{1}{4} - 3 a$
(v) Given $(1.1 m - 0.4) (1.1 m + 0.4) = (1.1 m)^{2} - (0.4)^{2}$
$∵ [(a + b) (a - b) = (a^{2} - b^{2})]$
$∴ (1.1 m - 0.4) (1.1 m + 0.4) = 1.21 m^{2} - 0.16$

(vi) Given $(a^{2} + b^{2}) (- a^{2} + b^{2})$ $= (b^{2} + a^{2}) (b^{2} - a^{2})$
$= (b^{2})^{2} - (a^{2})^{2}$ $∵$ $[(a + b) (a - b) = (a^{2} - b^{2})]$
$∴ (a^{2} + b^{2}) (- a^{2} + b^{2}) = b^{4} - a^{4}$

(vii) Given $(6 x - 7) (6 x + 7) = (6 x)^{2} - (7)^{2}$ $∵ [(a + b) (a - b) = (a^{2} - b^{2})]$
$∴ (6 x - 7) (6 x + 7) = 36 x^{2} - 49$

(viii) $(- a + c) (- a + c)$ $= (- a + c)^{2}$
$= (- a)^{2} + c^{2} - 2 (- a) (c)$ $∵ [(a - b)^{2}$ $= a^{2} + b^{2} - 2 a b]$
$∴ (- a + c) (- a + c) = a^{2} - 2 a c + c^{2}$
(ix) Given $(\frac{x}{2} + \frac{3 y}{4}) (\frac{x}{2} + \frac{3 y}{4})$
$= {(\begin{matrix} \frac{x}{2} + \frac{3 y}{4} \end{matrix})}^{2}$
$= {(\begin{matrix} \frac{x}{2} \end{matrix})}^{2} + 2 (\begin{matrix} \frac{x}{2} \end{matrix}) (\begin{matrix} \frac{3 y}{4} \end{matrix}) + {(\begin{matrix} \frac{3 y}{4} \end{matrix})}^{2}$
$= \frac{x^{2}}{4} + \frac{3 x y}{4} + \frac{9 y^{2}}{16}$ $∵ [(a + b)^{2}$ $= a^{2} + b^{2} + 2 a b]$

(x) Given $(7 a - 9 b) (7 a - 9 b) = (7 a - 9 b)^{2}$
$= (7 a)^{2} + (9 b)^{2} - 2 (7 a) (9 b)$ $∵ [(a - b)^{2}$ $= a^{2} + b^{2} - 2 a b]$
$= 49 a^{2} + 81 b^{2} - 126 a b$

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Similar questions

Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a − 7) (2a − 7) (iv)

(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a² + b²) (− a² + b²)

(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)

(ix) (x) (7a − 9b) (7a − 9b)

Q. Use the identity

(x + a) (x + b) = x^{2} + (a + b) x + a b

to find following products.
(i)

(x + 3) (x + 7)

(4 x + 5) (4 x + 1)

(4 x - 5) (4 x - 1)

(4 x + 5) (4 x - 1)

(2 x + 5 y) (2 x + 3 y)

(2 a^{2} + 9) (2 a^{2} + 5)

(x y z - 4) (x y z - 2)

Q. Expand each of the following, using suitable identities:
(i)

(x + 2 y + 4 z)^{2}

(2 x - y + z)^{2}

(- 2 x + 3 y + 2 z)^{2}

(3 a - 7 b - c)^{2}

(- 2 x + 5 y - 3 z)^{2}

{[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}

Q. Question 4
Expand each of the following, using suitable identities:
(i)

(x + 2 y + 4 z)^{2}

(2 x - y + z)^{2}

(- 2 x + 3 y + 2 z)^{2}

(3 a - 7 b - c)^{2}

(- 2 x + 5 y - 3 z)^{2}

{[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}

Q. Question 1(ix)
Use a suitable identity to get each of the following products:

(\frac{x}{2} + \frac{3 y}{4}) (\frac{x}{2} + \frac{3 y}{4})

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