What are the solutions to the equation (p+1)(p+3)=0 ?

The solutions to the equation (p+1)(p+3)=0 are p=-1,p=-3

Find the zeros of (p+1)(p+3)=0

The zeros of (p+1)(p+3)=0 are p=-1,p=-3

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(p + 1) (p + 3) = 0

p = - 1, p = - 3

Solution steps

(p + 1) (p + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

p + 1 = 0 or p + 3 = 0

Solve

p + 1 = 0 : p = - 1

Solve

p + 3 = 0 : p = - 3

The solutions to the quadratic equation are:

p = - 1, p = - 3

12 x^{2} = 17 x + 5

\frac{16 i ^{2}}{2 i}

(x - 2)^{4}

15 h^{2} - 17 h - 4

4 x^{2} + 36 x + 32

What are the solutions to the equation (p+1)(p+3)=0 ?
The solutions to the equation (p+1)(p+3)=0 are p=-1,p=-3
Find the zeros of (p+1)(p+3)=0
The zeros of (p+1)(p+3)=0 are p=-1,p=-3