What are the solutions to the equation 12x^4-7x^2-6x+16x^3=0 ?

The solutions to the equation 12x^4-7x^2-6x+16x^3=0 are x=0,x=-1/2 ,x= 2/3 ,x=-3/2

Find the zeros of 12x^4-7x^2-6x+16x^3=0

The zeros of 12x^4-7x^2-6x+16x^3=0 are x=0,x=-1/2 ,x= 2/3 ,x=-3/2

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12 x^{4} - 7 x^{2} - 6 x + 16 x^{3} = 0

x = 0, x = - \frac{1}{2}, x = \frac{2}{3}, x = - \frac{3}{2}

Decimal Notation

x = 0, x = - 0.5, x = 0.66666 \dots, x = - 1.5

Solution steps

12 x^{4} - 7 x^{2} - 6 x + 16 x^{3} = 0

Factor

12 x^{4} - 7 x^{2} - 6 x + 16 x^{3} : x (2 x + 1) (3 x - 2) (2 x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

Solve

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

The solutions are

x = 0, x = - \frac{1}{2}, x = \frac{2}{3}, x = - \frac{3}{2}

19 = 32 (1 - b)^{26}

x^{3} + x = - 2

x^{3} + x = 10

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

n^{3} = 60

What are the solutions to the equation 12x^4-7x^2-6x+16x^3=0 ?
The solutions to the equation 12x^4-7x^2-6x+16x^3=0 are x=0,x=-1/2 ,x= 2/3 ,x=-3/2
Find the zeros of 12x^4-7x^2-6x+16x^3=0
The zeros of 12x^4-7x^2-6x+16x^3=0 are x=0,x=-1/2 ,x= 2/3 ,x=-3/2