Find $p$ if $(x + 3)$ is a factor of $x^3 - x^2 + px + 15$.

Question

I'm just making sure I answered this correctly.

If $(x+3)$ is a factor, then $P(-3)$ would equal $0$, correct?

Precisely, so compute $P(-3) = 0$ and you will get the value of $p$, — Andreas Caranti, May 20 '13 at 09:14

score 6 · Accepted Answer · answered May 20 '13 at 09:14

6

Yes, that's correct. If $(x+3)$ is a factor, then $P(-3) = 0$ by the Factor Theorem. So

\[P(-3) = (-3)^3-(-3)^2-3p+15 = -27-9-3p+15 = -3p-21 = 0 ,\] and so $p = -7$.

answered May 20 '13 at 09:14

Wisław

I'll give it a best answer once the website allows me :P – missiledragon May 20 '13 at 09:17

score 2 · Answer 2 · answered May 20 '13 at 09:17

Yes, if $(x+3)$ is a factor of $x^3-x^2+px+15$, it means that you can write :

$$x^3-x^2+px+15 = Q(x)(x+3),\text{where } Q(x) \text{ is a polynomial of degree 2}$$

So when $x=-3$, we have $(-3)^3-(-3)^2+p(-3)+15 = Q(-3)(-3+3) = 0$

In order to find $p$, you have to do the euclidian division and see for what $p$ the remainder of this division equals $0$.

score 0 · Answer 3 · answered May 20 '13 at 15:00

0

Alternatively: $\,-3\,$ root of $\ x^3+27-(x^2-p\,x -4(-3))\ \Rightarrow\ p = -4 -3 $

answered May 20 '13 at 15:00

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