What are the solutions to the equation (2x-8)(x+4)=0 ?

The solutions to the equation (2x-8)(x+4)=0 are x=4,x=-4

Find the zeros of (2x-8)(x+4)=0

The zeros of (2x-8)(x+4)=0 are x=4,x=-4

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(2 x - 8) (x + 4) = 0

x = 4, x = - 4

Solution steps

(2 x - 8) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 8 = 0 or x + 4 = 0

Solve

2 x - 8 = 0 : x = 4

Solve

x + 4 = 0 : x = - 4

The solutions to the quadratic equation are:

x = 4, x = - 4

x, 5 x^{2} + 6 x y + 5 y^{2} - 4 x + 4 y = 0

2 x^{2} + 4 x = 2

x^{2} + x = 0

(x^{2} - 10 x + 25) + 6 = 42

\frac{1}{2} x^{2} = - 32

What are the solutions to the equation (2x-8)(x+4)=0 ?
The solutions to the equation (2x-8)(x+4)=0 are x=4,x=-4
Find the zeros of (2x-8)(x+4)=0
The zeros of (2x-8)(x+4)=0 are x=4,x=-4