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Question

The factors of 12ab + 9a + 12 + 16b are


A

(4b+3)(3a+4)

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B

(4a+3)(3b+4)

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C

(2b+3)(3a+6)

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D

(4b-3)(3a-4)

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Solution

The correct option is A

(4b+3)(3a+4)


12ab + 9a + 12 + 16b = 3a × (4b + 3) + 4 × (3 + 4b)

The factors of 12ab + 9a + 12 + 16b are (4b + 3) × (3a + 4).

So, 4b + 3 and 3a + 4 are the irreducible factors of the given algebraic expression.


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