Factorize- 16 a - b 2-4 a -4 bA- 4a-b4 a-4 b-1B- 4a-b4 a-4 b-1C- 16 a - b a - b -1D- 4 a - b 4 a -4 b -1

The correct option is D 4(a+b)(4a+4b 1)⇒ 16(a+b)^2 4a 4b ⇒ 16(a+b)^2 4(a + b) Take out 4(a + b) common, we get, ⇒ 4(a + b)[4(a+b) 1] ⇒ 4(a + b)(4a ...

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Byju's Answer

Standard IX

Mathematics

Taking Out Common Factors

Factorize: 16...

Question

Factorize: $16 (a + b)^{2} - 4 a - 4 b$

4 (a + b) (4 a + 4 b + 1)

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4 (a + b) (4 a - 4 b - 1)

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16 (a + b) (a + b - 1)

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4 (a + b) (4 a + 4 b - 1)

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Solution

The correct option is D $4 (a + b) (4 a + 4 b - 1)$
$\Rightarrow 16 (a + b)^{2} - 4 a - 4 b$

$\Rightarrow 16 (a + b)^{2} - 4 (a + b)$

$Take out 4 (a + b) common, we get,$

$\Rightarrow 4 (a + b) [4 (a + b) - 1]$

$\Rightarrow 4 (a + b) (4 a + 4 b - 1)$

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Similar questions

Factorize by the grouping method :

16 (a + b)^{2} - 4 a - 4 b

Q. Factorize the following equation by the grouping method

16 {(a + b)}^{2} - 4 a - 4 b

Factorise by grouping method:

$16 (a + b)^{2} - 4 a - 4 b$

Q. Factorize :

a - b - 4 a^{2} + 4 b^{2}

Question 2 (ix)

Add:

ab - 4a, 4b - ab, 4a - 4b

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Factorize: 16(a+b)2−4a−4b

The correct option is D 4(a+b)(4a+4b−1)⇒16(a+b)2−4a−4b ⇒16(a+b)2−4(a+b) Take out 4(a+b) common, we get, ⇒4(a+b)[4(a+b)−1] ⇒4(a+b)(4a+4b−1)

Factorize: $16 (a + b)^{2} - 4 a - 4 b$

The correct option is D $4 (a + b) (4 a + 4 b - 1)$
$\Rightarrow 16 (a + b)^{2} - 4 a - 4 b$

$\Rightarrow 16 (a + b)^{2} - 4 (a + b)$

$Take out 4 (a + b) common, we get,$

$\Rightarrow 4 (a + b) [4 (a + b) - 1]$

$\Rightarrow 4 (a + b) (4 a + 4 b - 1)$