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}=(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)