What are the solutions to the equation 2x^3+3x^2-32x+15=0 ?

The solutions to the equation 2x^3+3x^2-32x+15=0 are x=3,x= 1/2 ,x=-5

Find the zeros of 2x^3+3x^2-32x+15=0

The zeros of 2x^3+3x^2-32x+15=0 are x=3,x= 1/2 ,x=-5

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2 x^{3} + 3 x^{2} - 32 x + 15 = 0

x = 3, x = \frac{1}{2}, x = - 5

Decimal Notation

x = 3, x = 0.5, x = - 5

Solution steps

2 x^{3} + 3 x^{2} - 32 x + 15 = 0

Factor

2 x^{3} + 3 x^{2} - 32 x + 15 : (x - 3) (2 x - 1) (x + 5)

(x - 3) (2 x - 1) (x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or 2 x - 1 = 0 or x + 5 = 0

Solve

x - 3 = 0 : x = 3

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

x + 5 = 0 : x = - 5

The solutions are

x = 3, x = \frac{1}{2}, x = - 5

3 x^{3} - 2 x^{2} - 82 x - 10 = 0

2 x^{4} - 12 x^{2} - 54 = 0

2 x^{3} - 24 x = 0

0 = x^{3} + 2 x^{2} - 9 x - 18

π \cdot r^{3} = \frac{1}{4}

What are the solutions to the equation 2x^3+3x^2-32x+15=0 ?
The solutions to the equation 2x^3+3x^2-32x+15=0 are x=3,x= 1/2 ,x=-5
Find the zeros of 2x^3+3x^2-32x+15=0
The zeros of 2x^3+3x^2-32x+15=0 are x=3,x= 1/2 ,x=-5