What are the solutions to the equation (6x^2+11x+4)(2x-3b)=0 ?

The solutions to the equation (6x^2+11x+4)(2x-3b)=0 are x=-1/2 ,x=-4/3 ,x=(3b)/2

Find the zeros of (6x^2+11x+4)(2x-3b)=0

The zeros of (6x^2+11x+4)(2x-3b)=0 are x=-1/2 ,x=-4/3 ,x=(3b)/2

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(6 x^{2} + 11 x + 4) (2 x - 3 b) = 0

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

Solution steps

(6 x^{2} + 11 x + 4) (2 x - 3 b) = 0

Factor

(6 x^{2} + 11 x + 4) (2 x - 3 b) : (2 x + 1) (3 x + 4) (2 x - 3 b)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

Solve

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

The solutions are

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

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

b^{3} = - 216

18 x^{4} - 3 x^{2} - 1 = 0

(z + 1)^{5} = z^{5}

y, x = y^{3} - 1

What are the solutions to the equation (6x^2+11x+4)(2x-3b)=0 ?
The solutions to the equation (6x^2+11x+4)(2x-3b)=0 are x=-1/2 ,x=-4/3 ,x=(3b)/2
Find the zeros of (6x^2+11x+4)(2x-3b)=0
The zeros of (6x^2+11x+4)(2x-3b)=0 are x=-1/2 ,x=-4/3 ,x=(3b)/2