Question

-6x^3 -7x^2 +qx - 12 is divisible by 3x-4. There is no remainder. Find the value of q.

Answer 1

If
`3x-4` is a factor of
`p(x)=-6x^3-7x^2+qx-12`, then
`x=\frac43` is a root of
`p(x)` so that
`p\left(\frac43\right)=0`:

`-6\left(\frac43\right)^3-7\left(\frac43\right)^2+q\left(\frac43\right)-12=0`

`\frac{4q-116}3=0`

`4q-116=0`

`4(q-29)=0`

`\implies\boxed{q=29}`