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

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

Find the zeros of 3x^3+4x^2=12x+16

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

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3 x^{3} + 4 x^{2} = 12 x + 16

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

Decimal

x = - 1.33333 \dots, x = - 2, x = 2

Solution steps

3 x^{3} + 4 x^{2} = 12 x + 16

Move

16

to the left side

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

Move

12 x

to the left side

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

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

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

r^{3} + 2 r^{2} - 5 r - 6 = 0

2 x^{4} - x^{3} + x^{2} - x - 1

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

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

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