What are the solutions to the equation x^3+4x^2-29x+24=0 ?

The solutions to the equation x^3+4x^2-29x+24=0 are x=1,x=3,x=-8

Find the zeros of x^3+4x^2-29x+24=0

The zeros of x^3+4x^2-29x+24=0 are x=1,x=3,x=-8

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x^{3} + 4 x^{2} - 29 x + 24 = 0

x = 1, x = 3, x = - 8

Solution steps

x^{3} + 4 x^{2} - 29 x + 24 = 0

Factor

x^{3} + 4 x^{2} - 29 x + 24 : (x - 1) (x - 3) (x + 8)

(x - 1) (x - 3) (x + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 3 = 0 or x + 8 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

Solve

x + 8 = 0 : x = - 8

The solutions are

x = 1, x = 3, x = - 8

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

(x + 2) (5 x^{2} + 9) (x^{2} + 3) = 0

2 x^{3} - 18 x^{2} - 8 x + 72 = 0

(x + 7) (x^{2} - 49) = 0

16 x^{4} + 9 = 25 x^{2}

What are the solutions to the equation x^3+4x^2-29x+24=0 ?
The solutions to the equation x^3+4x^2-29x+24=0 are x=1,x=3,x=-8
Find the zeros of x^3+4x^2-29x+24=0
The zeros of x^3+4x^2-29x+24=0 are x=1,x=3,x=-8