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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!!!!

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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 $x-r$)! :)

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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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