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There's a math question on an online test which asks the following

Multiply the following expression, and simplify: $\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$

But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I can't seem to figure out how to simplify it enough to get it right.

$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$ equals, $\frac{x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2-4xy}$. I then factored x out of the numerator and denominator to get $\frac{x(x^3 + 4x^2y - 16xy^2 - 64y^3)}{x(x-4y)}$ and cancelled out the factored x's to get $\frac{x^3 + 4x^2y - 16xy^2 - 64y^3}{x-4y}$. I don't know what to do from here though.

I've managed to get enough marks to be able to pass it but since it's a readiness test I want to understand all of the material going in.

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  • $\begingroup$ I just thought of trying polynomial long division but I'm heading out for a family dinner in a few minutes. Is that applicable? $\endgroup$
    – Greener
    Dec 24, 2013 at 21:20
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    $\begingroup$ the numerator of the second fraction can be factored to $x(x+4y)$. The $x$ then cancels with the one in the denominator of the first fraction. $\endgroup$
    – K. Rmth
    Dec 24, 2013 at 21:22
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    $\begingroup$ I accidentally wrote the question down wrong initially. $\endgroup$
    – Greener
    Dec 25, 2013 at 5:43

5 Answers 5

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It looks like you are applying fraction addition rules to fraction multiplication:

$$\frac ab+\frac cd=\frac {a\cdot d+b\cdot c}{b\cdot d}$$

Instead you should use:

$$\frac ab\cdot \frac cd=\frac {a\cdot c}{b\cdot d}$$

So we have

$$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac {(x^2+16y^2)(x+4y)}{x-4y}=\frac {x^3+4x^2y+16xy^2+64y^3}{x-4y}$$

Of course, it could be that there was a negative sign mixup which lead to the result

$$\frac {x^3+4x^2y-16xy^2-64y^3}{x-4y}=\frac {(x^2-16y^2)(x+4y)}{x-4y}=(x+4y)^2$$

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  • $\begingroup$ How did you get $(x+4y)^2$ from $\frac {(x^2-16y^2)(x+4y)}{x-4y}$? I did make a mistake on the negative sign for $\frac{x^2+16y^2}{x}$. It should be $\frac{x^2-16y^2}{x}$ $\endgroup$
    – Greener
    Dec 25, 2013 at 5:48
  • $\begingroup$ Nevermind, a below post clarified it. $\endgroup$
    – Greener
    Dec 25, 2013 at 6:15
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Let's start from the very beginning here. Your expression is $$\frac {x^2 - 16y^2}{x} \cdot \frac {x^2 + 4xy}{x - 4y}$$ You can multiply the numerators and denominators to get this: $$\frac {x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2 - 4xy}$$ When you divide via the polynomial long division method, you get $$x^2 + 8xy + 16y^2$$ When you factor, you get: $$(x + 4y)^2$$

So, your answer is $(x + 4y)^2$.

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The cancellation make sense if instead of $x^2+16y^2$ we use $x^2-16y^2$ $$\frac{x^2-16y^2}{x}\frac{x^2+4xy}{x-4y}=\frac{x^2-(4y)^2}{x}\frac{x(x+4y)}{x-4y}=$$ $$=\frac{(x-4y)(x+4y)}{x}\frac{x(x+4y)}{x-4y}=(x+4y)^2$$

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$$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=(x^2+16y^2)\cdot \frac{(x+4y)}{x-4y}$$

$$=(x+i4y)(x-i4y) \frac{x+4y}{x-4y}$$

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As mentioned in the comments, by factoring out $x$ in the second fraction, $x^2+4xy$ becomes $x(x+4y)$. Which leaves you with:

$$\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=\frac{(x^2+16y^2)(x+4y)}{x-4y}$$

Alternately if you had not seen the cancellations and had ended up with $x^3+16xy^2+4x^2y+64y^3$ as your numerator. Grouping $x^3+4x^2y$ and $16xy^2+64y^3$ together, notice that $x^2$ and $16y^2$ are common (respectively). Factor them out to get, $$x^2(x+4y) + 16y^2(x+4y) = (x^2+16y^2)(x+4y)$$

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