Question

How do you find all the zeros of `f(x) = x⁴ - 10x² + 24`?

Answer 1

`x = +-2" "` and `" "x = +-sqrt(6)`

Explanation:

Given:

`f(x) = x^4-10x^2+24`

Note that this quartic contains only terms of even degree, so we can start to factor it as a quadratic in `x^2`.

Note also that `10 = 4 + 6` and `24 = 4 * 6`

Hence we find:

`x^4-10x^2+24 = (x^2-4)(x^2-6)`

`color(white)(x^4-10x^2+24) = (x^2-2^2)(x^2-(sqrt(6))^2)`

`color(white)(x^4-10x^2+24) = (x-2)(x+2)(x-sqrt(6))(x+sqrt(6))`

Hence zeros:

`x = +-2" "` and `" "x = +-sqrt(6)`

graph{x^4-10x^2+24 [-5.067, 4.933, -6, 32]}

Answer 2

`x=\pm2, \pm\sqrt6`

Explanation:

Given function:

`f(x)=x^4-10x^2+24`

The zeroes of above bi-quadratic polynomial is given by setting `f(x)=0` as follows

`x^4-10x^2+24=0`

`x^4-6x^2-4x^2+24=0`

`x^2(x^2-6)-4(x^2-6)=0`

`(x^2-6)(x^2-4)=0`

`x^2-6=0\ \ or\ \ x^2-4=0`

`x^2=6\ \ or\ \ x^2=4`

`x=\pm\sqrt6\ \ or\ \ x=\pm 2`

`x=\pm2, \pm\sqrt6`