Question

This is the question

Answer 1

`=\left(9a-8\right)\left(a+1\right)`

Explanation:

`9a^2+a-8\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(9a^2-8a\right)+\left(9a-8\right)\\\mathrm{Factor\:out\:}a\mathrm{\:from\:}9a^2-8a\mathrm{:\quad }a\left(9a-8\right)\\9a^2-8a\\\mathrm{Apply\:exponent\:rule}:\quad \:a^(b+c)=a^ba^c\\a^2=aa\\=9aa-8a\\\mathrm{Factor\:out\:common\:term\:}a\\=a\left(9a-8\right)\\=a\left(9a-8\right)+\left(9a-8\right)\\\mathrm{Factor\:out\:common\:term\:}9a-8\\=\left(9a-8\right)\left(a+1\right)`

Answer 2

So the right answer is (9a-8)(a+1)

Look at the attached picture

Hope it will help you