Question

$\text{Which of the following are the factors of}12ab+9a+12+16b?$

• A. $(3a-4)\text{and}(4b-3)$
• B. $(3a-4)\text{and}(4b+3)$
• C. $(3a+4)\text{and}(4b+3)$
• D. $(3a+4)(4b-3)$

Answer 1

The correct option is C $(3a+4)\text{and}(4b+3)$

$\text{Given, the expression is}12ab+9a+12+16b.$

$\text{Taking 3a common from the first}\text{two terms and 4 common from the}\text{last two terms, we get}12ab+9a+12+16b=3a(4b+3)+4(3+4b)=(3a+4)(3+4b)$

$\text{So,}(4b+3)\text{and}(3a+4)\text{are the irreducible factors of the given}\text{expression.}$