What are the solutions to the equation (x^2-x)^2-14(x^2-x)+24=0 ?

The solutions to the equation (x^2-x)^2-14(x^2-x)+24=0 are x=-1,x=2,x=-3,x=4

Find the zeros of (x^2-x)^2-14(x^2-x)+24=0

The zeros of (x^2-x)^2-14(x^2-x)+24=0 are x=-1,x=2,x=-3,x=4

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(x^{2} - x)^{2} - 14 (x^{2} - x) + 24 = 0

x = - 1, x = 2, x = - 3, x = 4

Solution steps

(x^{2} - x)^{2} - 14 (x^{2} - x) + 24 = 0

Expand

(x^{2} - x)^{2} - 14 (x^{2} - x) + 24 : x^{4} - 2 x^{3} - 13 x^{2} + 14 x + 24

x^{4} - 2 x^{3} - 13 x^{2} + 14 x + 24 = 0

Factor

x^{4} - 2 x^{3} - 13 x^{2} + 14 x + 24 : (x + 1) (x - 2) (x + 3) (x - 4)

(x + 1) (x - 2) (x + 3) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x + 3 = 0 or x - 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x + 3 = 0 : x = - 3

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 1, x = 2, x = - 3, x = 4

0 = x^{4} - 25

x^{3} + 3 x^{2} - 6 x - 18 = 0

8 y^{3} - 27 = 0

x^{3} - 8 x^{2} - x + 8 = 0

3 x^{3} - 9 x^{2} + 3 x - 9 = 0

What are the solutions to the equation (x^2-x)^2-14(x^2-x)+24=0 ?
The solutions to the equation (x^2-x)^2-14(x^2-x)+24=0 are x=-1,x=2,x=-3,x=4
Find the zeros of (x^2-x)^2-14(x^2-x)+24=0
The zeros of (x^2-x)^2-14(x^2-x)+24=0 are x=-1,x=2,x=-3,x=4