What are the solutions to the equation (x^2+2x)(x^2+2x-7)=8 ?

The solutions to the equation (x^2+2x)(x^2+2x-7)=8 are x=-1,x=2,x=-4

Find the zeros of (x^2+2x)(x^2+2x-7)=8

The zeros of (x^2+2x)(x^2+2x-7)=8 are x=-1,x=2,x=-4

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(x^{2} + 2 x) (x^{2} + 2 x - 7) = 8

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

Solution steps

(x^{2} + 2 x) (x^{2} + 2 x - 7) = 8

Expand

(x^{2} + 2 x) (x^{2} + 2 x - 7) : x^{4} + 4 x^{3} - 3 x^{2} - 14 x

x^{4} + 4 x^{3} - 3 x^{2} - 14 x = 8

Move

8

to the left side

x^{4} + 4 x^{3} - 3 x^{2} - 14 x - 8 = 0

Factor

x^{4} + 4 x^{3} - 3 x^{2} - 14 x - 8 : (x + 1)^{2} (x - 2) (x + 4)

(x + 1)^{2} (x - 2) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x + 4 = 0 : x = - 4

The solutions are

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

- x^{4} + 7 x^{3} - 6 x^{2} = 0

x^{4} + 5 x^{2} - 14 = 0

x, x^{3} - y = 2

x, x^{3} - y = 1

2 x^{3} - x + 1 = 0

What are the solutions to the equation (x^2+2x)(x^2+2x-7)=8 ?
The solutions to the equation (x^2+2x)(x^2+2x-7)=8 are x=-1,x=2,x=-4
Find the zeros of (x^2+2x)(x^2+2x-7)=8
The zeros of (x^2+2x)(x^2+2x-7)=8 are x=-1,x=2,x=-4