What are the solutions to the equation x^3+4x^2-8x=32 ?

The solutions to the equation x^3+4x^2-8x=32 are x=-4,x=-2sqrt(2),x=2sqrt(2)

Find the zeros of x^3+4x^2-8x=32

The zeros of x^3+4x^2-8x=32 are x=-4,x=-2sqrt(2),x=2sqrt(2)

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x^{3} + 4 x^{2} - 8 x = 32

x = - 4, x = - 22, x = 22

Decimal Notation

x = - 4, x = - 2.82842 \dots, x = 2.82842 \dots

Solution steps

x^{3} + 4 x^{2} - 8 x = 32

Move

32

to the left side

x^{3} + 4 x^{2} - 8 x - 32 = 0

Factor

x^{3} + 4 x^{2} - 8 x - 32 : (x + 4) (x + 22) (x - 22)

(x + 4) (x + 22) (x - 22) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 22 = 0 or x - 22 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 22 = 0 : x = - 22

Solve

x - 22 = 0 : x = 22

The solutions are

x = - 4, x = - 22, x = 22

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

2 x^{3} - 5 x^{2} + x + 2 = 0

x^{4} - 13 x^{2} + 22 = 0

x^{3} + 11 x^{2} + 10 x = 0

4 t^{3} - 8 t = 0

What are the solutions to the equation x^3+4x^2-8x=32 ?
The solutions to the equation x^3+4x^2-8x=32 are x=-4,x=-2sqrt(2),x=2sqrt(2)
Find the zeros of x^3+4x^2-8x=32
The zeros of x^3+4x^2-8x=32 are x=-4,x=-2sqrt(2),x=2sqrt(2)