What are the solutions to the equation -x^3+x^2+8x-12=0 ?

The solutions to the equation -x^3+x^2+8x-12=0 are x=2,x=-3

Find the zeros of -x^3+x^2+8x-12=0

The zeros of -x^3+x^2+8x-12=0 are x=2,x=-3

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- x^{3} + x^{2} + 8 x - 12 = 0

x = 2, x = - 3

Solution steps

- x^{3} + x^{2} + 8 x - 12 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 3 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 3 = 0 : x = - 3

The solutions are

x = 2, x = - 3

5 x^{3} - 3 x^{2} + 10 x - 6 = 0

x^{3} - 12 x^{2} + 36 x + 44 = 0

- 2 (5 x + 6) (x^{2} - 64) = 0

x^{4} + 3 = 0

(2^{24})^{\frac{1}{2}} = p^{4}

What are the solutions to the equation -x^3+x^2+8x-12=0 ?
The solutions to the equation -x^3+x^2+8x-12=0 are x=2,x=-3
Find the zeros of -x^3+x^2+8x-12=0
The zeros of -x^3+x^2+8x-12=0 are x=2,x=-3