Question

adgdfghfdhdf

Answer 1

You're familiar with the rational root test correct?

Answer 2

yes

Answer 3

Great. So in this case, that tells you that any rational root must be of the form \(\pm d\) where \(d\) divides \(p\). Since \(p\) is a prime number, that tells you that \(d\) is either \(1\) or \(p\). Following so far?

Answer 4

yup

Answer 5

Now we just plug them in.\[f(1)=1+1+1-p=3-p\implies p=3\]\[f(-1)=(-1)^3+(-1)^2-1+p=-1+1-1+p=p-1\implies p=1\]\[f(p)=p^3+p^2+2p=p(p^2+p+2)=0\implies p=0,p^*\]\[f(-p)=-p^3+p^2=p^2(1-p)=0\implies p=0,1\]where \[p^*=\frac{-1\pm\sqrt{-7}}{2}.\]Only one of these actually happens when \(p\) is prime.

Answer 6

Still making sense?

Answer 7

yes

Answer 8

Hold on... I think I made a few sign errors (that are important). Give me a second to retype everything correctly.

Answer 9

ok

Answer 10

\(f(1)\) remains unchanged.\[f(-1)=-1+1-1-p=-1-p\implies p=-1\]\[f(p)=p^3+p^2+p-p=p^3+p^2=p^2(p+1)\implies p=0,-1\]\[f(-p)=-p^3+p^2-p-p=-p(p^2-p+2)\implies p=0,p^*\]where\[p^*=\frac{1\pm\sqrt{-7}}{2}.\]Sorry about that. But we still have the same result, that \(p\) must be 3.

Answer 11

ok. and p=3 is the only answer?

Answer 12

Yes. \(p=3\) should be the only possibility.

So the only polynomial is \(x^3+x^2+x-3\), which has a root \(1\). So now you can use polynomial long division or synthetic division to factor an \(x-1\) out.

Answer 13

What's your preferred method of factoring?

Answer 14

synthetic

Answer 15

Alright. Why don't you try that yourself, and I can check what you get at the end.

Answer 16

ok thanks

Answer 17

i got x^2+2x+3 as my answer

Answer 18

That looks great. So the polynomial factors as\[(x-1)(x^2+2x+3)\]Now does \(x^2+2x+3\) have any real roots?

Answer 19

nope

Answer 20

Correct. So that means the only real root is \(1\), and we've finished the problem.

Answer 21

ok thanks

Answer 22

You're welcome.