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I am not sure how I would factor this. The $x^4$ and $x^2$ are really throwing me off. Can someone explain how I would factor this?

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Let $y=x^2$, then factor the expression in $y$. – Joel Reyes Noche Apr 19 '13 at 4:33
I've heard this method colloquially referred to as 'chunking'. It works in other situations which might throw you off, e.g. $e^{2x}+ae^x+b$ becomes $y^2+ay+b$ with $y=e^x$. Solving the quadratic equation in $y$ and substituting back in is far easier than any alternative. – Ian Coley Apr 19 '13 at 4:35
Even if you don't recognize immediately that you can substitute $y=x^2$, you can work to that as follows. Note that the polynomial is even in $x$: replace $x$ with $-x$ and the polynomial stays the same. So, if $a$ is a root, then $-a$ is a root. So, it factors to $(x-a)(x+a)(x-b)(x+b)$ for some complex $a$ and $b$. Collect related factors to get $(x^2 - a^2)(x^2 - b^2)$. Give $a^2$ and $b^2$ simpler names, say $c$ and $d$, where these are possibly complex. It should then be clear. – Eric Jablow Apr 19 '13 at 5:23

Since all of the powers of $x$ in this polynomial are even ($18$ counts as $18 \cdot x^0$), you would make a substitution of $ t = x^2 $ . Since $x^4 = (x^2)^2$ , you can write your polynomial as $t^2 - 7t - 18$ . How would you factor that?

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Let $y=x^2$. You then get $y^2-7y-18$. Can you factor it now?

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Solution 1. \begin{eqnarray*} x^4-7x^2-18&=&(x^4+2x^2)-(9x^2+18)\\ &=&x^2(x^2+2)-9(x^2+2)\\ &=&(x^2+2)(x^2-9)\\ &=&(x^2+2)(x-3)(x+3) \end{eqnarray*}

Solution 2. \begin{eqnarray*} x^4-7x^2-18&=&(x^4-9x^2)+(2x^2-18)\\ &=&x^2(x^2-9)+2(x^2-9)\\ &=&(x^2-9)(x^2+2)\\ &=&(x-3)(x+3)(x^2+2) \end{eqnarray*}

Solution 3. \begin{eqnarray*} x^4-7x^2-18&=&(x^4-81)-(7x^2-63)\\ &=&(x^2+9)(x^2-9)-7(x^2-9)\\ &=&(x^2-9)(x^2+9-7)\\ &=&(x-3)(x+3)(x^2+2) \end{eqnarray*}

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Hint: For this one, note that $x$ only appears as an even power. Substitute $y$ for $x^2$ and see if you can do it.

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Hint: If the $x^4$ and $x^2$ are confusing, a very useful trick is to replace them.

More precisely, if we let "$y$" mean $x^2$, then the polynomial is $$y^2-7y-18.$$ Can you factor this? After you have done that, you can replace $y$ with $x^2$ and keep going.

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As most other people have commented, the most sensible thing to do is probably make the substitution $y=x^2$. In this way, $$ x^4-7x^2-18\tag{1} $$ becomes $$ y^2-7y-18\tag{2} $$ We can then factor $(2)$ as follows: $y^2-7y-18 = (y-9)(y+2)$. Since $y=x^2$, we see that $(1)$ factors as follows: $$ x^4-7x^2-18=(x^2-9)(x^2+2)=(x-3)(x+3)(x^2+2). $$ This is probably the most straightforward, easy way of going about it.

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