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

3 Answers
Apr 16, 2018

Answer:

#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)#

Apr 16, 2018

Answer:

#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)#

Apr 16, 2018

Answer:

#(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)#