What are the solutions to the equation (x+2b)(3x^2-16x+5)=0 ?

The solutions to the equation (x+2b)(3x^2-16x+5)=0 are x=-2b,x= 1/3 ,x=5

Find the zeros of (x+2b)(3x^2-16x+5)=0

The zeros of (x+2b)(3x^2-16x+5)=0 are x=-2b,x= 1/3 ,x=5

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(x + 2 b) (3 x^{2} - 16 x + 5) = 0

x = - 2 b, x = \frac{1}{3}, x = 5

Solution steps

(x + 2 b) (3 x^{2} - 16 x + 5) = 0

Factor

(x + 2 b) (3 x^{2} - 16 x + 5) : (x + 2 b) (3 x - 1) (x - 5)

(x + 2 b) (3 x - 1) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 b = 0 or 3 x - 1 = 0 or x - 5 = 0

Solve

x + 2 b = 0 : x = - 2 b

Solve

3 x - 1 = 0 : x = \frac{1}{3}

Solve

x - 5 = 0 : x = 5

The solutions are

x = - 2 b, x = \frac{1}{3}, x = 5

z^{3} + 4 = 5 + i

x^{3} - 3 x^{2} + x + 1 = 0

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

x^{3} + x^{2} - x + 15 = 0

x^{3} - 12 x^{2} + 47 x - 60 = 0

What are the solutions to the equation (x+2b)(3x^2-16x+5)=0 ?
The solutions to the equation (x+2b)(3x^2-16x+5)=0 are x=-2b,x= 1/3 ,x=5
Find the zeros of (x+2b)(3x^2-16x+5)=0
The zeros of (x+2b)(3x^2-16x+5)=0 are x=-2b,x= 1/3 ,x=5