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3 years ago

Is the binomial a factor of the polynomial function? f(x)=x3+2x2−16x−32

2 answers:

8 0

Since it is x^3 cubed it has 3 maximum roots
by the zero multiplication property of function
if you plug find a root then you can deduce it is a factor in other words if f(x)=(x-a)(x-b)
then f(a)=0 and f(b)=0
so
f(2)=8+8-32-32= -48 therefore it is not a factor
f(-2)=-8+8+32-32=0 hence it is a factor
f(4)=64+32-64-32=0 it is a factor
f(-4)=-64+32+64-32=0 it is a factor
f(-6) it is not a factor via the fundmental therom of algebra
f(x)=(x+2)(x-4)(x+4)
=(x^2-16)(x+2)=x^3+2x^2-16x-32 which checks our work

umka21 [38]3 years ago

3 0

Answer: (A) No, (B) Yes, (C) Yes, (D) Yes, (E) No.

Step-by-step explanation: We are given to whether each binomial is a factor of the following polynomial or not :

$f(x)=x^3+2x^2-16x-32.$

To select the correct factors, we need to factorize the given polynomial.

We will be using the following factorization formula :

$a^2-b^2=(a+b)(a-b).$

We have

$f(x)\\\\=x^3+2x^2-16x-32\\\\=x^2(x+2)-16(x+2)\\\\=(x+2)(x^2-16)\\\\=(x+2)(x^2-4^2)\\\\=(x+2)(x+4)(x-4).$

Therefore, the factors of the given polynomial are (x + 2), (x + 4) and (x - 4).

Thus,

option (A) : (x - 2)

No.

option (B) : (x + 2)

Yes.

option (C) : (x - 4)

Yes.

option (D) : (x + 4)

Yes.

option (A) : (x + 6)

No.

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