In my online course I also have to determine the remainder of this equation and I am stuck on it as we have never divided a polynomial by anything larger than (x-a) variables.

Determine the remainder when $x^3 + 3x^2 – x – 2$ is divided by $(x + 3)(x+ 5) $

I am not sure how to begin answering this question and am hoping someone can help me get started! That would be extremely appreciated!! Thank you so much for all the help!!!!


You need to use polynomial long division (try googling "polynomial long division") to divide $x^3+3x^2-x-2$ by $(x+3)(x+5)= x^2+8x+15$.

This is done by dividing leading terms, multiplying the resulting quotient by $x^2+8x+15$, subtracting this result from $x^3+3x^2-x-2$, and repeating...

The first step: the leading terms are $x^3$ and $x^2$. So $x^3/x^2 = x$. Now $x(x^2+8x+15) = x^3+8x^2+15x$. Subtracting this from $x^3+3x^2-x-2$, we get $-5x^2-16x-2$ (a remainder so far).

Now start over with $-5x^2-16x-2$ and $x^2+8x+15$. Dividing leading terms: $-5x^2/x^2 = -5$. Then $-5(x^2+8x+15) = -5x^2-40x-75$. Subtract this from $-5x^2-16x-2$ and get $24x+73$ (our remainder).

We cannot continue division any more since $24x+73$ has a lower degree than $x^2+8x+15$.

Therefore, $x^3+3x^2-x-2$ divided by $x^2+8x+15$ is $x-5$ (the terms we found along the way: $x$ and $-5$) with a remainder of $24x+73$.

Or we can use technology: Wolfram Alpha

Another way to find the remainder is to use the fact that it must be linear (at least one degree lower than $x^2+8x+15=(x+3)(x+5)$. So division should result in $x^3+3x^2-x-2 = q(x)(x+3)(x+5) + (ax+b)$ where $q(x)$ is the quotient and $ax+b$ is the remainder.

If we plug in $x=-3$, we get: $(-3)^3+3(-3)^2-(-3)-2 = q(-3)(-3+3)(-3+5) + (a(-3)+b)$. So $1=-3a+b$. Likewise if we plug in $x=-5$ we get: $(-5)^3+3(-5)^2-(-5)-2 = q(-5)(-5+3)(-5+5)+(a(-5)+b)$. So $-47=-5a+b$.

Subtracting $-47=-5a+b$ from $1=-3a+b$ yields $48=2a$ so $a=24$. Then $1=-3(24)+b$ so $b=73$.

Just like before the remainder is $ax+b = 24x+73$.

If you divide a polynomial $f(x)$ by a linear factor $(x-r)$, this same (second) technique says: $f(x)=q(x)(x-r)+???$ so $f(r)=q(r)(r-r)+???$ so $f(r)=???$. In other words, the remainder when dividing by a linear factor is just $f(r)$ (the polynomial evaluated at the root of $x-r$)! :)


Here's the process that Bill Cook mentioned, latexed nicely. Hope this helps. \begin{align}&\phantom{=\overline{)}} x\phantom{^3}- \phantom{00}5 \\ x^2 + 8x + 15 \text{ }&\phantom{=}\overline{) \text{ } x^3 + \phantom{1}3x^2 -\phantom{15} x - 2} \\ &\phantom{=\overline{)}\text{ }}\underline{ x^3 + \phantom{1}8x^2 +15x}\phantom{-}\downarrow\\ &\phantom{=\overline{)}\text{ }x^3} - \phantom{1}5 x^2-16x-2 \\ &\phantom{=\overline{)}\text{ }X^1}\underline{-\phantom{1}5x^2 - 40x- 75} \\ &\phantom{=\overline{)}\text{ }x^3+\text{}\phantom{1}3x^2}+24x + 73 \end{align}


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