What are the solutions to the equation (x+4)(x+1)=0 ?

The solutions to the equation (x+4)(x+1)=0 are x=-4,x=-1

Find the zeros of (x+4)(x+1)=0

The zeros of (x+4)(x+1)=0 are x=-4,x=-1

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(x + 4) (x + 1) = 0

x = - 4, x = - 1

Solution steps

(x + 4) (x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 1 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 1 = 0 : x = - 1

The solutions to the quadratic equation are:

x = - 4, x = - 1

5^{3}

\frac{11}{5} \cdot \frac{7}{2}

\frac{2 π}{\frac{8 π}{3}}

\frac{2 k}{3 k - 7} - \frac{3}{8 k}

x - 1 > x - 3

What are the solutions to the equation (x+4)(x+1)=0 ?
The solutions to the equation (x+4)(x+1)=0 are x=-4,x=-1
Find the zeros of (x+4)(x+1)=0
The zeros of (x+4)(x+1)=0 are x=-4,x=-1