What are the solutions to the equation 16x^3+96x^2-x-6=0 ?

The solutions to the equation 16x^3+96x^2-x-6=0 are x=-6,x=-1/4 ,x= 1/4

Find the zeros of 16x^3+96x^2-x-6=0

The zeros of 16x^3+96x^2-x-6=0 are x=-6,x=-1/4 ,x= 1/4

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16 x^{3} + 96 x^{2} - x - 6 = 0

x = - 6, x = - \frac{1}{4}, x = \frac{1}{4}

Decimal Notation

x = - 6, x = - 0.25, x = 0.25

Solution steps

16 x^{3} + 96 x^{2} - x - 6 = 0

Factor

16 x^{3} + 96 x^{2} - x - 6 : (x + 6) (4 x + 1) (4 x - 1)

(x + 6) (4 x + 1) (4 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 6 = 0 or 4 x + 1 = 0 or 4 x - 1 = 0

Solve

x + 6 = 0 : x = - 6

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

4 x - 1 = 0 : x = \frac{1}{4}

The solutions are

x = - 6, x = - \frac{1}{4}, x = \frac{1}{4}

x^{4} + x^{3} + 4 x + 4 = 0

20 = x^{3} + 4 x^{2} - 5 x

x^{3} - 343 = 0

(x - 1)^{3} + 8 = 0

8 x^{3} - 14 x^{2} + 7 x - 1 = 0

What are the solutions to the equation 16x^3+96x^2-x-6=0 ?
The solutions to the equation 16x^3+96x^2-x-6=0 are x=-6,x=-1/4 ,x= 1/4
Find the zeros of 16x^3+96x^2-x-6=0
The zeros of 16x^3+96x^2-x-6=0 are x=-6,x=-1/4 ,x= 1/4