Question

Factorise $x^4-5x^2+4$ using factor theorem

• A. f(x)=(x-1)(x+1)(x-2)(x+2)
• B. f(x)=(x-1)(x+1)(x-3)(x+3)
• C. f(x)=(x-2)(x+2)(x-3)(x+3)
• D. f(x)=(x-5)(x+5)(x-3)(x+3 )

Answer 1

The correct option is A f(x)=(x-1)(x+1)(x-2)(x+2)

Let $f\left(x\right)=x^4-5x^2+4$
factors of 4 = $\pm 1,\pm 2,\pm 4$
$f\left(1\right)=(1{)}^4-5(1{)}^2+4=1-5+4=0$
$f(-1)=(-1{)}^4-5(-1{)}^2+4=1-5+4=0$
$f\left(2\right)=(2{)}^4-5(2{)}^2+4=16-20+4$
$\therefore f\left(2\right)=0$
$f(-2)=(-2{)}^4-5(-2{)}^2+4=16-20+4$
$\therefore f(-2)=0$
$f\left(4\right)=(4{)}^4-5(4{)}^2+4=256-80+4$
$=180\ne 0$
$f(-4)=(-4{)}^4-5(-4{)}^2+4=180\ne 0$

By fator theorem,
(x-1), (x+1), (x-2) and (x+2) are factors of f(x).
So, f(x)=(x-1)(x+1)(x-2)(x+2)

Thus, the correct answer is option (a).