Question

3a^2 -3a - 36. solving quadratic by factoring. factor each expression. be sure to check for greatest common factor first.

Answer 1

we have the expression

`3a^2-3a-36`

step 1

Factor 3

`3(a^2-a-12)`

step 2

equate to zero

`3(a^2-a-12)=0`

step 3

Solve

`(a^2-a-12)=0`
`\begin{gathered} a^2-a=12 \\ (a^2-a+(1)/(4)-(1)/(4))=12 \\ (a^2-a+(1)/(4))=12+(1)/(4) \\ (a^2-a+(1)/(4))=(49)/(4) \end{gathered}`

Rewrite as perfect squares

`(a-(1)/(2))^2=(49)/(4)`

take the square root on both sides

`\begin{gathered} a-(1)/(2)=\pm(7)/(2) \\ a=(1)/(2)\pm(7)/(2) \end{gathered}`

the values of a are

a=4 and a=-3

therefore

`3(a^2-a-12)=3(a-4)(a+3)`