Question

If $16{\left(4a-3b\right)}^2+4\left(4a-3b\right)$ is factorised, then its factors are

• A. $4\left(4a-3b\right)\left(16a-12b-4\right)$
• B. $4\left(4a-3b\right)\left(16a+12b+4\right)$
• C. $4\left(4a+3b\right)\left(16a-12b+4\right)$
• D. $4\left(4a-3b\right)\left(16a-12b+1\right)$

Answer 1

The correct option is D $4\left(4a-3b\right)\left(16a-12b+1\right)$

$16{(4a-3b)}^2+4(4a-3b)=4(4a-3b)[4(4a-3b)+1]=4(4a-3b)[16a-12b+1]$