SOLUTION: which of the following polynomials has a factor of x-1?. A) {{{p(x)=x^3+x^2-2x+1}}} B) {{{q(x)=2x^3-x^2+x-1}}} c) {{{r(x)=3x^3-x-2}}} D) {{{s(x)=-3x^3+3x+1}}}

Algebra ->  Expressions -> SOLUTION: which of the following polynomials has a factor of x-1?. A) {{{p(x)=x^3+x^2-2x+1}}} B) {{{q(x)=2x^3-x^2+x-1}}} c) {{{r(x)=3x^3-x-2}}} D) {{{s(x)=-3x^3+3x+1}}}      Log On


   



Question 1040787: which of the following polynomials has a factor of x-1?.
A) p%28x%29=x%5E3%2Bx%5E2-2x%2B1
B) q%28x%29=2x%5E3-x%5E2%2Bx-1
c) r%28x%29=3x%5E3-x-2
D) s%28x%29=-3x%5E3%2B3x%2B1

Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
By the remainder theorem, if x-k is factor of p(x), then p(k) = 0.

In this case, x-k = x-1, so k = 1. Therefore, if we find a function p(x) such that p(k) = p(1) = 0 then we have found the answer.

Essentially we plug in x = 1 into each answer choice. Then we see which results in 0.

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Let's go through the answer choices one by one. Starting with choice A

p%28x%29+=+x%5E3%2Bx%5E2-2x%2B1
p%281%29+=+1%5E3%2B1%5E2-2%281%29%2B1 Replace every x with 1
p%281%29+=+1%2B1-2%281%29%2B1
p%281%29+=+1%2B1-2%2B1
p%281%29+=+1

Since the result is NOT equal to 0, this means that x-1 is NOT a factor of p(x).

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Now onto choice B. Like before, plug in x = 1 to see if we get a result of 0.

q%28x%29+=+2x%5E3-x%5E2%2Bx-1
q%281%29+=+2%281%29%5E3-%281%29%5E2%2B1-1 Replace every 'x' with 1
q%281%29+=+2%281%29-1%2B1-1
q%281%29+=+2-1%2B1-1
q%281%29+=+1

Again we don't get 0. So choice B is not the answer either.
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Part C)

r%28x%29+=+3x%5E3-x-2
r%281%29+=+3%281%29%5E3-1-2 Plug in x = 1
r%281%29+=+3%281%29-1-2
r%281%29+=+3-1-2
r%281%29+=+0

We got a result of 0, so x-1 is a factor of r(x). We have our answer. Let's check part D just to complete the problem.
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Part D)

s%28x%29+=+-3x%5E3%2B3x%2B1
s%281%29+=+-3%281%29%5E3%2B3%281%29%2B1 Plug in x = 1
s%281%29+=+-3%281%29%2B3%281%29%2B1
s%281%29+=+-3%2B3%2B1
s%281%29+=+1

The nonzero result means x-1 is not a factor of s(x)
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To summarize, the final answer is Choice C

Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.
which of the following polynomials has a factor of x-1?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Those and only those that have the number "1" as their root.

(based on the "Remainder theorem", see the lesson Divisibility of polynomial f(x) by binomial x-a in this site).

A) p%28x%29 = x%5E3%2Bx%5E2-2x%2B1

B) q%28x%29 = 2x%5E3-x%5E2%2Bx-1

c) r%28x%29 = 3x%5E3-x-2         <--- This and only this

D) s%28x%29 = -3x%5E3%2B3x%2B1