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lesantik [10]
3 years ago
7

PLEASE HELP ME please!

Mathematics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

x < -1

so its C

Step-by-step explanation:

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If x^2 +x-12 is a factor of x^3+ax^2-10x-b then find the values of a and b
Marysya12 [62]

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

7 0
4 years ago
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