What are the solutions to the equation (x^2+x)^2-18(x^2+x)+72=0 ?

The solutions to the equation (x^2+x)^2-18(x^2+x)+72=0 are x=2,x=-4,x=-3,x=3

Find the zeros of (x^2+x)^2-18(x^2+x)+72=0

The zeros of (x^2+x)^2-18(x^2+x)+72=0 are x=2,x=-4,x=-3,x=3

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(x^{2} + x)^{2} - 18 (x^{2} + x) + 72 = 0

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

Solution steps

(x^{2} + x)^{2} - 18 (x^{2} + x) + 72 = 0

Expand

(x^{2} + x)^{2} - 18 (x^{2} + x) + 72 : x^{4} + 2 x^{3} - 17 x^{2} - 18 x + 72

x^{4} + 2 x^{3} - 17 x^{2} - 18 x + 72 = 0

Factor

x^{4} + 2 x^{3} - 17 x^{2} - 18 x + 72 : (x - 2) (x + 4) (x + 3) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x + 4 = 0 : x = - 4

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

The solutions are

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

(w - 3) (w^{2} + 3 w + 9) = 2

x^{6} - 5 x^{4} + 6 x^{2} - 1 = 0

- x^{3} - 11 x^{2} - 39 x - 45 = 0

x^{4} + 6 x^{2} + 6 x^{3} + 6 x + 5 = 0

m^{3} - 6 m^{2} + 12 m - 8 = 0

What are the solutions to the equation (x^2+x)^2-18(x^2+x)+72=0 ?
The solutions to the equation (x^2+x)^2-18(x^2+x)+72=0 are x=2,x=-4,x=-3,x=3
Find the zeros of (x^2+x)^2-18(x^2+x)+72=0
The zeros of (x^2+x)^2-18(x^2+x)+72=0 are x=2,x=-4,x=-3,x=3