Is it possible to factor y=x^4 - 13x^2 + 36y=x4−13x2+36? If so, what are the factors?

3 Answers
sjc
Apr 16, 2018

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

Explanation:

x^4-13x+36x4−13x+36

u=x^2u=x2

u^2-13u+36u2−13u+36

a quadratic in u

=(u-4)(u-9)=(u−4)(u−9)

(x^2-4)(x^2-9)(x2−4)(x2−9)

both brackets are differences of squares

(x+2)(x-2)(x+3)(x-3)(x+2)(x−2)(x+3)(x−3)

iceman
Apr 16, 2018

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

Explanation:

y=x^4-13x^2+36y=x4−13x2+36 => Let: x^2 = ux2=u, then: x^4=u^2x4=u2:
y=u^2-13u+36y=u2−13u+36 => find 2 no's that multiply to 36, add to -13:
y=u^2-9u-4u+36y=u2−9u−4u+36
y=u(u-9)-4(u-9)y=u(u−9)−4(u−9)
y=(u-9)(u-4)y=(u−9)(u−4) => substitute back x^2=ux2=u:
y=(x^2-9)(x^2-4)y=(x2−9)(x2−4) => factor using the difference of squares:
y=(x-3)(x+3)(x-2)(x+2)y=(x−3)(x+3)(x−2)(x+2)

Jim G.
Apr 16, 2018

(x+3)(x-3)(x-2)(x+2)(x+3)(x−3)(x−2)(x+2)

Explanation:

"using the "color(blue)"substitution "u=x^2using the substitution u=x2

rArrx^4-13x^2+36=u^2-13u+36⇒x4−13x2+36=u2−13u+36

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

rArru^2-13u+36=(u-9)(u-4)⇒u2−13u+36=(u−9)(u−4)

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

(x^2-9)(x^2-4)(x2−9)(x2−4)

"both "x^2-9" and "x^2-4" are "color(blue)"difference of squares"both x2−9 and x2−4 are difference of squares

"which factorise, in general as"which factorise, in general as

•color(white)(x)a^2-b^2=(a-b)(a+b)∙xa2−b2=(a−b)(a+b)

rArrx^2-9=(x-3)(x+3)⇒x2−9=(x−3)(x+3)

rArrx^2-4=(x-2)(x+2)⇒x2−4=(x−2)(x+2)

rArrx^4-13x^2+36=(x-3)(x+3)(x-2)(x+2)⇒x4−13x2+36=(x−3)(x+3)(x−2)(x+2)

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