What are the solutions to the equation x^4-4x^3-x^2+16x-12=0 ?

The solutions to the equation x^4-4x^3-x^2+16x-12=0 are x=1,x=3,x=-2,x=2

Find the zeros of x^4-4x^3-x^2+16x-12=0

The zeros of x^4-4x^3-x^2+16x-12=0 are x=1,x=3,x=-2,x=2

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x^{4} - 4 x^{3} - x^{2} + 16 x - 12 = 0

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

Solution steps

x^{4} - 4 x^{3} - x^{2} + 16 x - 12 = 0

Factor

x^{4} - 4 x^{3} - x^{2} + 16 x - 12 : (x - 1) (x - 3) (x + 2) (x - 2)

(x - 1) (x - 3) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x - 3 = 0 : x = 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

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

x^{3} + x^{2} - 9 x - 9

x^{3} - 16 x^{2} + 75 x - 90 = 0

x, \frac{d ^{2} x}{d t ^{2}} = 2 x^{3}

x^{4} + 11 x^{2} = 0

x^{3} - x = 0

What are the solutions to the equation x^4-4x^3-x^2+16x-12=0 ?
The solutions to the equation x^4-4x^3-x^2+16x-12=0 are x=1,x=3,x=-2,x=2
Find the zeros of x^4-4x^3-x^2+16x-12=0
The zeros of x^4-4x^3-x^2+16x-12=0 are x=1,x=3,x=-2,x=2