What are the solutions to the equation (x-5)(x^2+4x+4)=8(x+2) ?

The solutions to the equation (x-5)(x^2+4x+4)=8(x+2) are x=-2,x=-3,x=6

Find the zeros of (x-5)(x^2+4x+4)=8(x+2)

The zeros of (x-5)(x^2+4x+4)=8(x+2) are x=-2,x=-3,x=6

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(x - 5) (x^{2} + 4 x + 4) = 8 (x + 2)

x = - 2, x = - 3, x = 6

Solution steps

(x - 5) (x^{2} + 4 x + 4) = 8 (x + 2)

Move

8 (x + 2)

to the left side

(x - 5) (x^{2} + 4 x + 4) - 8 (x + 2) = 0

Factor

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

(x + 2) (x + 3) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x + 3 = 0 or x - 6 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x + 3 = 0 : x = - 3

Solve

x - 6 = 0 : x = 6

The solutions are

x = - 2, x = - 3, x = 6

3 x^{3} - 19 x^{2} + 33 x - 9 = 0

6 x^{3} + 25 x^{2} - 24 x + 5

y^{4} - 4 y^{2} + 3 = 0

x^{3} + 5 x + 5 = 0

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

What are the solutions to the equation (x-5)(x^2+4x+4)=8(x+2) ?
The solutions to the equation (x-5)(x^2+4x+4)=8(x+2) are x=-2,x=-3,x=6
Find the zeros of (x-5)(x^2+4x+4)=8(x+2)
The zeros of (x-5)(x^2+4x+4)=8(x+2) are x=-2,x=-3,x=6