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The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks

4 Answers 4

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Your assumptions are that there are polynomials $q,s,t$ such that $$ \begin{cases} f(x) = (x-2)(x+3) q(x) &+ a x + b\\ f(x) = (x -2) s(x) &+ 5\\ f(x) = (x + 3) t(x) \end{cases} $$

Now calculate $f(2)$ and $f(-3)$.

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    I can only find out that f(2)=5 and f(-3)=0 by this. How can I find a and b? – 2017-01-10
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    Plug first $x = 2$ and then $x = -3$ in the first equation to get other expressions for $f(2)$ and $f(-3)$. Equate to the expressions you already have. – 2017-01-10
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    But then I still have q(x) unknown – 2017-01-10
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    Nobody asked you to determine $q(x)$, which is an impossible task. You are only looking for $a$ and $b$. Why don't you substitute $x = 2$ in the first equation and see what you get, ***please***? – 2017-01-10
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    I finally understand this now thank you so much! – 2017-01-10
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    @Estudiantekriss, you're welcome. If you like my answer, please consider upvoting/accepting it. – 2017-01-10
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Simple: first write $\;f(x)=(x+3)g(x)$, and perform the Euclidean division of $g(x)$ by $x-2$: $$g(x)=(x-2)q(x)+g(2).$$ Now $f(2)=5g(2)$, so that $$f(x)=(x-2)(x+3)q(x)+ g(2)(x+3)=(x-2)(x+3)q(x)+ \underbrace{\frac{f(2)}5(x+3)}_{\text{remainder}}. $$

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Yes you are on the right track.

Let f(x) = (x-2)(x+3) x m + (ax + b).

We know f(2) = 5, so (2-2)(2+3) x m + (2 a + b) = 5 And similarly f(-3) = 0, so (-3-2)(-3+3) x m + (-3 a + b) = 0

This gives 2 equations

(2 a + b) = 5, and

(-3 a + b) = 0

Which you can solve to give a=1 and b=3

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Hint:

If $$f(x)=(x-\alpha)g(x)+r$$ then $r=f(\alpha)$

Really, $x=\alpha$ $f(\alpha)=0+r$

Then $$f(x)=(x-2)(x+3)+ax+b$$

$f(2)=5$ and $f(-3)=0$

$$2a+b=5$$ $$-3a+b=0$$

$$a=1, b=3$$