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zvonat [6]
3 years ago
11

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

Mathematics
2 answers:
goblinko [34]3 years ago
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,

<em><u>option (A) : (x - 2)</u></em>

No.

<em><u>option (B) : (x + 2)</u></em>

Yes.

<em><u>option (C) : (x - 4)</u></em>

Yes.

<em><u>option (D) : (x + 4)</u></em>

Yes.

<em><u>option (A) : (x + 6)</u></em>

No.

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