What are the solutions to the equation (x+3)(x+12)=0 ?

The solutions to the equation (x+3)(x+12)=0 are x=-3,x=-12

Find the zeros of (x+3)(x+12)=0

The zeros of (x+3)(x+12)=0 are x=-3,x=-12

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(x + 3) (x + 12) = 0

x = - 3, x = - 12

Solution steps

(x + 3) (x + 12) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x + 12 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x + 12 = 0 : x = - 12

The solutions to the quadratic equation are:

x = - 3, x = - 12

n^{2} + 16 - 10 n

\frac{3 - 6 x}{3} = 7 x - 1

\frac{1}{4} - 25 z^{2}

\frac{x + 3}{2} + \frac{2 x - 5}{3} = \frac{1}{2}

(\frac{3}{8})^{- 2}

What are the solutions to the equation (x+3)(x+12)=0 ?
The solutions to the equation (x+3)(x+12)=0 are x=-3,x=-12
Find the zeros of (x+3)(x+12)=0
The zeros of (x+3)(x+12)=0 are x=-3,x=-12