Question

factor x^4-5x^2+4 completely

Answer 1

Set

$x^4-5x^2+4=0$

Then, if y=x^2;

$\begin{gathered} y=x^2 \\ \Rightarrow x^4-5x^2+4=y^2-5y+4=0 \\ \Rightarrow y^2-5y+4=0 \end{gathered}$

Solve for y using the quadratic equation formula, as shown below

$\begin{gathered} \Rightarrow y=\frac{-(-5)\pm\sqrt[]{(-5)^2-4\cdot1\cdot4}_{}}{2\cdot1}=\frac{5\pm\sqrt[]{25-16}}{2}=\frac{5\pm3}{2} \\ \Rightarrow y=4,1 \end{gathered}$

Therefore, since y=x^2,

$\begin{gathered} \Rightarrow x^2=4,x^2=1 \\ \Rightarrow x=\pm2,x=\pm1 \end{gathered}$

Finally,

$\Rightarrow x^4-5x^2+4=(x-2)(x+2)(x-1)(x+1)$

The factored form of x^4-5x^2+4 is (x-2)(x+2)(x-1)(x+1)