Is it possible to factor y=x^4 - 13x^2 + 36? If so, what are the factors?

Apr 16, 2018

$y = \left(x + 2\right) \left(x - 2\right) \left(x + 3\right) \left(x - 3\right)$

Explanation:

${x}^{4} - 13 x + 36$

$u = {x}^{2}$

${u}^{2} - 13 u + 36$

a quadratic in u

$= \left(u - 4\right) \left(u - 9\right)$

$\left({x}^{2} - 4\right) \left({x}^{2} - 9\right)$

both brackets are differences of squares

$\left(x + 2\right) \left(x - 2\right) \left(x + 3\right) \left(x - 3\right)$

Apr 16, 2018

$y = \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right)$

Explanation:

$y = {x}^{4} - 13 {x}^{2} + 36$ => Let: ${x}^{2} = u$, then: ${x}^{4} = {u}^{2}$:
$y = {u}^{2} - 13 u + 36$ => find 2 no's that multiply to 36, add to -13:
$y = {u}^{2} - 9 u - 4 u + 36$
$y = u \left(u - 9\right) - 4 \left(u - 9\right)$
$y = \left(u - 9\right) \left(u - 4\right)$ => substitute back ${x}^{2} = u$:
$y = \left({x}^{2} - 9\right) \left({x}^{2} - 4\right)$ => factor using the difference of squares:
$y = \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right)$

Apr 16, 2018

$\left(x + 3\right) \left(x - 3\right) \left(x - 2\right) \left(x + 2\right)$

Explanation:

$\text{using the "color(blue)"substitution } u = {x}^{2}$

$\Rightarrow {x}^{4} - 13 {x}^{2} + 36 = {u}^{2} - 13 u + 36$

$\text{the factors of + 36 which sum to - 13 are - 9 and - 4}$

$\Rightarrow {u}^{2} - 13 u + 36 = \left(u - 9\right) \left(u - 4\right)$

$\text{change u back into terms of x gives}$

$\left({x}^{2} - 9\right) \left({x}^{2} - 4\right)$

$\text{both "x^2-9" and "x^2-4" are "color(blue)"difference of squares}$

$\text{which factorise, in general as}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$\Rightarrow {x}^{2} - 9 = \left(x - 3\right) \left(x + 3\right)$

$\Rightarrow {x}^{2} - 4 = \left(x - 2\right) \left(x + 2\right)$

$\Rightarrow {x}^{4} - 13 {x}^{2} + 36 = \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right)$