How to factor $x^4-7x^2-18$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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?

Answer 5

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*}