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Factorization of Expressions in the Form of ax+by and a²-b² as a Product of Terms

Factorize by ...

Question

Factorize by the grouping method :
$16 (a + b)^{2} - 4 a - 4 b$

4 (a - 3 b) (4 a + 4 b - 1)

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4 (2 a + b) (a + 4 b - 3)

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

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

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Solution

The correct option is D $4 (a + b) (4 a + 4 b - 1)$
$16 (a + b)^{2} - 4 a - 4 b$
$= 16 (a + b)^{2} - 4 (a + b)$
$= 4 (a + b) [4 (a + b) - 1]$
$= 4 (a + b) [4 a + 4 b - 1]$

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Similar questions

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

Factorise by grouping method:

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

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

Question 2 (ix)

Add:

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

\frac{{cos}^{4} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

\frac{{cos}^{4} B}{{cos}^{2} A} + \frac{{sin}^{4} B}{{sin}^{2} A} = 1

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