What are the solutions to the equation 2x^3-3x^2-32x+48=0 ?

The solutions to the equation 2x^3-3x^2-32x+48=0 are x= 3/2 ,x=-4,x=4

Find the zeros of 2x^3-3x^2-32x+48=0

The zeros of 2x^3-3x^2-32x+48=0 are x= 3/2 ,x=-4,x=4

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2 x^{3} - 3 x^{2} - 32 x + 48 = 0

x = \frac{3}{2}, x = - 4, x = 4

Decimal Notation

x = 1.5, x = - 4, x = 4

Solution steps

2 x^{3} - 3 x^{2} - 32 x + 48 = 0

Factor

2 x^{3} - 3 x^{2} - 32 x + 48 : (2 x - 3) (x + 4) (x - 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x - 3 = 0 : x = \frac{3}{2}

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = \frac{3}{2}, x = - 4, x = 4

x^{5} = x

x^{5} - 2 x^{4} + 5 x^{3} - 8 x^{2} + 4 x = 0

8 x^{7} - 56 x^{6} = 0

x^{4} + 6 x^{3} + 18 x^{2} + 30 x + 25 = 0

z^{3} - 6 z - 4 = 0

What are the solutions to the equation 2x^3-3x^2-32x+48=0 ?
The solutions to the equation 2x^3-3x^2-32x+48=0 are x= 3/2 ,x=-4,x=4
Find the zeros of 2x^3-3x^2-32x+48=0
The zeros of 2x^3-3x^2-32x+48=0 are x= 3/2 ,x=-4,x=4