SOLVE
(x^2 – x)2 – 14(x^2 – x) + 24 = 0
A. 3, 1, -2, -4
B. 1, -3, -4, 2
C. -3, -1, 2, 4
D. -2, 3, -4, 1
just treat it as a quadratic:
u^2 - 14u + 24 = 0
(u-2)(u-12) = 0
so, now we have
x^2-x = 2
or
x^2-x = 12
Solving those, we have
x^2-x-2=0: x = -1,2
x^2-x-12=0: x = -3,4
So, we finally have
x = -3,-1,2,4
check:
(x+3)(x+1)(x-2)(x-4) = x^4-2x^3-13x^2+14x+24
(x^2–x)^2–14(x^2–x)+24 = x^4-2x^3-13x^2+14x+24