Question

{x}^{4} - 17x ^{2} + 16factor completely

Answer 1

We can make a u-substitution to make it into a quadratic.

Let

$u=x^2$

Performing this substitution, we have:

$\begin{gathered} x^4-17x^2+16 \\ =(x^2)^2-17(x^2)+16 \\ \text{Making u-substitution,} \\ u^2-17u+16 \\ (u-16)(u-1) \end{gathered}$

Now, taking it back to "x", we have:

$\begin{gathered} (u-16)(u-1) \\ (x^2-16)(x^2-1) \end{gathered}$

We can further break it down by using the rule:

$a^2-b^2=(a+b)(a-b)$

So, let's factor it out completely:

$\begin{gathered} (x^2-16)(x^2-1) \\ ((x)^2-(4)^2)((x)^2-(1)^2) \\ (x-4)(x+4)(x-1)(x+1) \end{gathered}$

Thus, the fully factored form is:

$(x-4)(x+4)(x-1)(x+1)$