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

Answer:

a=3

b=24

Step-by-step 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 $x^2[tex] is 1 so all we have to look for are two numbers that multiply to be -12 and add to be positive 1 which in this case is 4 and -3.-12=4(-3) while 1=4+(-3).So the factored form of [tex]x^2+x-12$ 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