Question

Factorise: x^4– 5x^2 + 4​

Answer 1

(x - 2)(x - 1)(x + 1)(x + 2)

Explanation:

`{x}^(4) - 5 {x}^(2) + 4 \\ \\ = {x}^(4) - 4 {x}^(2) - {x}^(2) + 4 \\ \\ = {x}^(2) ( {x}^(2) - 4) - 1( {x}^(2) - 4) \\ \\ = ( {x}^(2) - 4)( {x}^(2) - 1) \\ \\ = (x + 2)(x - 2)(x + 1)(x - 1) \\ \\ = (x - 2)(x - 1)(x + 1)(x + 2)`

Answer 2

`(x+1)(x-1)(x+2)(x-2)`

Explanation:

To factorise
`x^4-5x^2+4` :

Let
`x^2=u`

`\implies x^4-5x^2+4=u^2-5u+4`

Now factorise
`u^2-5u+4`

`\textsf{for} \ ax^2+bx+c \ \textsf{find} \ v, w \ \textsf{such that} \ v \cdot w=a \cdot c \ \textsf{and} \ v+w=b`

`\textsf{and group into} \ (ax^2+vx)+(wx+c)`

`\implies v=-1, w=-4`

`\implies u^2-u-4u+4`

Break into groups

`\implies (u^2-u)+(-4u+4)`

Factor each parentheses:

`\implies u(u-1)-4(u-1)`

Factor out common term
`(u-1)`:

`\implies (u-1)(u-4)`

Substitute back
`u=x^2` :

`\implies (x^2-1)(x^2-4)`

Factor both parentheses using the Difference of Two Squares:
`a^2-b^2=(a+b)(a-b)`

`\implies (x+1)(x-1)(x+2)(x-2)`

Therefore, the factorisation of
`x^4-5x^2+4` is

`(x+1)(x-1)(x+2)(x-2)`