How to factor $x^4-7x^2-18$ 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?
 A: 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?
A: 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*}
A: Let $y=x^2$.  You then get $y^2-7y-18$.  Can you factor it now?
A: 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.
A: 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.
A: 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. 
