Question

factorise: x^4– 5x^2 + 4​

Answer 1

Answer:

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

Step-by-step 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

Answer:

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

Step-by-step 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

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