What are the solutions to the equation x^4-9x^2+4x+12=0 ?

The solutions to the equation x^4-9x^2+4x+12=0 are x=-1,x=2,x=-3

Find the zeros of x^4-9x^2+4x+12=0

The zeros of x^4-9x^2+4x+12=0 are x=-1,x=2,x=-3

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x^{4} - 9 x^{2} + 4 x + 12 = 0

x = - 1, x = 2, x = - 3

Solution steps

x^{4} - 9 x^{2} + 4 x + 12 = 0

Factor

x^{4} - 9 x^{2} + 4 x + 12 : (x + 1) (x - 2)^{2} (x + 3)

(x + 1) (x - 2)^{2} (x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x + 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x + 3 = 0 : x = - 3

The solutions are

x = - 1, x = 2, x = - 3

0 = 2 x^{6} - 27 x^{4}

y^{5} - 12 y^{4} + 24 y^{3} = 0

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

s^{3} = 27

17 c^{1} + 5 c^{2} + 255 c^{3} = 53

What are the solutions to the equation x^4-9x^2+4x+12=0 ?
The solutions to the equation x^4-9x^2+4x+12=0 are x=-1,x=2,x=-3
Find the zeros of x^4-9x^2+4x+12=0
The zeros of x^4-9x^2+4x+12=0 are x=-1,x=2,x=-3