What are the solutions to the equation 48x^3+40x^2-x-3=0 ?

The solutions to the equation 48x^3+40x^2-x-3=0 are x=-1/3 ,x= 1/4 ,x=-3/4

Find the zeros of 48x^3+40x^2-x-3=0

The zeros of 48x^3+40x^2-x-3=0 are x=-1/3 ,x= 1/4 ,x=-3/4

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48 x^{3} + 40 x^{2} - x - 3 = 0

x = - \frac{1}{3}, x = \frac{1}{4}, x = - \frac{3}{4}

Decimal Notation

x = - 0.33333 \dots, x = 0.25, x = - 0.75

Solution steps

48 x^{3} + 40 x^{2} - x - 3 = 0

Factor

48 x^{3} + 40 x^{2} - x - 3 : (3 x + 1) (4 x - 1) (4 x + 3)

(3 x + 1) (4 x - 1) (4 x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or 4 x - 1 = 0 or 4 x + 3 = 0

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

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

Solve

4 x + 3 = 0 : x = - \frac{3}{4}

The solutions are

x = - \frac{1}{3}, x = \frac{1}{4}, x = - \frac{3}{4}

y^{9} = - 8

2 x^{3} + 14 x^{2} + 24 x = 0

x, f (x^{3}) = x^{2}

x = x^{4} - 1

x, x^{4} - 8 x^{2} + 7 = 0

What are the solutions to the equation 48x^3+40x^2-x-3=0 ?
The solutions to the equation 48x^3+40x^2-x-3=0 are x=-1/3 ,x= 1/4 ,x=-3/4
Find the zeros of 48x^3+40x^2-x-3=0
The zeros of 48x^3+40x^2-x-3=0 are x=-1/3 ,x= 1/4 ,x=-3/4