Question

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

Answer 1

`y=(x+2)(x-2)(x+3)(x-3)`

Explanation:

`x^4-13x+36`

`u=x^2`

`u^2-13u+36`

a quadratic in u

`=(u-4)(u-9)`

`(x^2-4)(x^2-9)`

both brackets are differences of squares

`(x+2)(x-2)(x+3)(x-3)`

Answer 2

`y=(x-3)(x+3)(x-2)(x+2)`

Explanation:

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

Answer 3

`(x+3)(x-3)(x-2)(x+2)`

Explanation:

`"using the "color(blue)"substitution "u=x^2`

`rArrx^4-13x^2+36=u^2-13u+36`

`"the factors of + 36 which sum to - 13 are - 9 and - 4"`

`rArru^2-13u+36=(u-9)(u-4)`

`"change u back into terms of x gives"`

`(x^2-9)(x^2-4)`

`"both "x^2-9" and "x^2-4" are "color(blue)"difference of squares"`

`"which factorise, in general as"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`rArrx^2-9=(x-3)(x+3)`

`rArrx^2-4=(x-2)(x+2)`

`rArrx^4-13x^2+36=(x-3)(x+3)(x-2)(x+2)`