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

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

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

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

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

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

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x - 1 = 0 : x = \frac{1}{4}

Solve

x + 3 = 0 : x = - 3

Solve

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

The solutions are

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

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

5 y^{3} + 2 y^{2} - 45 y - 18 = 0

(5 x - 2)^{3} + 10 = 18

10 x^{4} - 17 x^{2} + 3 = 0

x^{3} + 8 x^{2} - 4 x - 32 = 0

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