Question

If x^2 +x-12 is a factor of x^3+ax^2-10x-b then find the values of a and b

Answer 1

a=3

b=24

Explanation:

If
`x^2+x-12` is a factor of
`x^3+ax^2-10x-b`, then the factors of
`x^2+x-12` must also be factors of
`x^3+ax^2-10x-b`.

So what are the factors of
`x^2+x-12`? Well the cool thing here is the coefficient of
is
`(x+4)(x-3)`.

The zeros of
`x^2+x-12` are therefore x=-4 and x=3. We know those are zeros of
`x^2+x-12` by the factor theorem.

So x=-4 and x=3 are also zeros of
`x^3+ax^2-10x-b` because we were told that
`x^2+x-12` was a factor of it.

This means that when we plug in -4, the result will be 0. It also means when we plug in 3, the result will be 0.

Let's do that.

`(-4)^3+a(-4)^2-10(-4)-b=0` Equation 1.

`(3)^3+a(3)^2-10(3)-b=0` Equation 2.

Let's simplify Equation 1 a little bit:

`(-4)^3+a(-4)^2-10(-4)-b=0`

`-64+16a+40-b=0`

`-24+16a-b=0`

`16a-b=24`

Let's simplify Equation 2 a little bit:

`(3)^3+a(3)^2-10(3)-b=0`

`27+9a-30-b=0`

`-3+9a-b=0`

`9a-b=3`

So we have a system of equations to solve:

16a-b=24

9a-b=3

---------- This is setup for elimination because the b's are the same. Let's subtract the equations.

16a-b=24

9a-b= 3

------------------Subtracting now!

7a =21

Divide both sides by 7:

a =3

Now use one the equations with a=3 to find b.

How about 9a-b=3 with a=3.

So plug in 3 for a.

9a-b=3

9(3)-b=3

27-b=3

Subtract 27 on both sides:

-b=-24

Multiply both sides by -1:

b=24

So a=3 and b=24