JRAHS 3U 2007 Trial Poly Question (1 Viewer)

APOLLO2609 · Mar 22, 2019

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).

The answer is:

P(x) = 16x-x³

Thanks

fan96 · Mar 22, 2019

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

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

JRAHS 3U 2007 Trial Poly Question (1 Viewer)

APOLLO2609

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fan96

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