What are the solutions to the equation -x^3-4x^2+19x-14=0 ?

The solutions to the equation -x^3-4x^2+19x-14=0 are x=1,x=2,x=-7

Find the zeros of -x^3-4x^2+19x-14=0

The zeros of -x^3-4x^2+19x-14=0 are x=1,x=2,x=-7

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- x^{3} - 4 x^{2} + 19 x - 14 = 0

x = 1, x = 2, x = - 7

Solution steps

- x^{3} - 4 x^{2} + 19 x - 14 = 0

Factor

- x^{3} - 4 x^{2} + 19 x - 14 : - (x - 1) (x - 2) (x + 7)

- (x - 1) (x - 2) (x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 2 = 0 or x + 7 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

Solve

x + 7 = 0 : x = - 7

The solutions are

x = 1, x = 2, x = - 7

6 x^{3} + x^{2} - 4 x + 1 = 0

x^{3} - 2.1 x^{2} + 0.21 = 0

x^{3} - 121 x = 0

x^{3} + x = x^{2} + 1

2 x^{3} + 3 x^{2} - 36 x + 6 = 0

What are the solutions to the equation -x^3-4x^2+19x-14=0 ?
The solutions to the equation -x^3-4x^2+19x-14=0 are x=1,x=2,x=-7
Find the zeros of -x^3-4x^2+19x-14=0
The zeros of -x^3-4x^2+19x-14=0 are x=1,x=2,x=-7