Respuesta :
Answer:
a=3
b=24
Step-by-step explanation:
If [tex]x^2+x-12[/tex] is a factor of [tex]x^3+ax^2-10x-b[/tex], then the factors of [tex]x^2+x-12[/tex] must also be factors of [tex]x^3+ax^2-10x-b[/tex].
So what are the factors of [tex]x^2+x-12[/tex]? Well the cool thing here is the coefficient of [tex]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[/tex] is [tex](x+4)(x-3)[/tex].
The zeros of [tex]x^2+x-12[/tex] are therefore x=-4 and x=3. We know those are zeros of [tex]x^2+x-12[/tex] by the factor theorem.
So x=-4 and x=3 are also zeros of [tex]x^3+ax^2-10x-b[/tex] because we were told that [tex]x^2+x-12[/tex] 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.
[tex](-4)^3+a(-4)^2-10(-4)-b=0[/tex] Equation 1.
[tex](3)^3+a(3)^2-10(3)-b=0[/tex] Equation 2.
Let's simplify Equation 1 a little bit:
[tex](-4)^3+a(-4)^2-10(-4)-b=0[/tex]
[tex]-64+16a+40-b=0[/tex]
[tex]-24+16a-b=0[/tex]
[tex]16a-b=24[/tex]
Let's simplify Equation 2 a little bit:
[tex](3)^3+a(3)^2-10(3)-b=0[/tex]
[tex]27+9a-30-b=0[/tex]
[tex]-3+9a-b=0[/tex]
[tex]9a-b=3[/tex]
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
