# Thread: JRAHS 3U 2007 Trial Poly Question

1. ## JRAHS 3U 2007 Trial Poly Question

Hey, I need help with this question. An explanation would be much appreciated as there are solutions (which I don't understand). The question is:

P(x) is an odd polynomial of degree 3. It has (x+4) as a factor and, when it is divided by (x-3), the remainder is 21. Find P(x).

P(x) = 16x-x3

Thanks

2. ## Re: JRAHS 3U 2007 Trial Poly Question

Let $P(x) = ax^3+bx^2+cx+d$.

Because $P$ is odd, $P(x) = -P(-x)$ and therefore $b = d = 0$.

$P(x) = ax^3+cx$.

By the remainder theorem, $P(-4) = 0$ and $P(3) = 21$.

Performing substitutions, we obtain

$\begin{cases}16a + c &= 0 \\ 9a + c &= 7 \end{cases} \implies a= -1,\, c =16$

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•