What are the solutions to the equation (x^2+3x+2)^2-8(x^2+3x)=4 ?

The solutions to the equation (x^2+3x+2)^2-8(x^2+3x)=4 are x=0,x=1,x=-3,x=-4

Find the zeros of (x^2+3x+2)^2-8(x^2+3x)=4

The zeros of (x^2+3x+2)^2-8(x^2+3x)=4 are x=0,x=1,x=-3,x=-4

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(x^{2} + 3 x + 2)^{2} - 8 (x^{2} + 3 x) = 4

x = 0, x = 1, x = - 3, x = - 4

Solution steps

(x^{2} + 3 x + 2)^{2} - 8 (x^{2} + 3 x) = 4

Expand

(x^{2} + 3 x + 2)^{2} - 8 (x^{2} + 3 x) : x^{4} + 6 x^{3} + 5 x^{2} - 12 x + 4

x^{4} + 6 x^{3} + 5 x^{2} - 12 x + 4 = 4

Move

4

to the left side

x^{4} + 6 x^{3} + 5 x^{2} - 12 x = 0

Factor

x^{4} + 6 x^{3} + 5 x^{2} - 12 x : x (x - 1) (x + 3) (x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 0, x = 1, x = - 3, x = - 4

x^{6} + 72 x^{3} + 512 = 0

x^{3} - 4 x^{2} = 0

x^{3} = 3 x

4 x - 4 x^{3} = 0

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

What are the solutions to the equation (x^2+3x+2)^2-8(x^2+3x)=4 ?
The solutions to the equation (x^2+3x+2)^2-8(x^2+3x)=4 are x=0,x=1,x=-3,x=-4
Find the zeros of (x^2+3x+2)^2-8(x^2+3x)=4
The zeros of (x^2+3x+2)^2-8(x^2+3x)=4 are x=0,x=1,x=-3,x=-4