What are the solutions to the equation 2x^3+3x^2-17x+12=0 ?

The solutions to the equation 2x^3+3x^2-17x+12=0 are x=1,x= 3/2 ,x=-4

Find the zeros of 2x^3+3x^2-17x+12=0

The zeros of 2x^3+3x^2-17x+12=0 are x=1,x= 3/2 ,x=-4

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2 x^{3} + 3 x^{2} - 17 x + 12 = 0

x = 1, x = \frac{3}{2}, x = - 4

Solution steps

2 x^{3} + 3 x^{2} - 17 x + 12 = 0

Factor

2 x^{3} + 3 x^{2} - 17 x + 12 : (x - 1) (2 x - 3) (x + 4)

(x - 1) (2 x - 3) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 2 x - 3 = 0 or x + 4 = 0

Solve

x - 1 = 0 : x = 1

Solve

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

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 1, x = \frac{3}{2}, x = - 4

z^{4} - 49 = 0

0.000004 x^{3} - 400 = 0

2 x (64 x^{2}) - 8 x^{2} (8 x) = 1

x^{2} + 2 x^{8} = 0

x^{3} = 2^{- 3}

What are the solutions to the equation 2x^3+3x^2-17x+12=0 ?
The solutions to the equation 2x^3+3x^2-17x+12=0 are x=1,x= 3/2 ,x=-4
Find the zeros of 2x^3+3x^2-17x+12=0
The zeros of 2x^3+3x^2-17x+12=0 are x=1,x= 3/2 ,x=-4