What are the solutions to the equation (3+2p)(2p-3)(p+1)=0 ?

The solutions to the equation (3+2p)(2p-3)(p+1)=0 are p=-3/2 ,p= 3/2 ,p=-1

Find the zeros of (3+2p)(2p-3)(p+1)=0

The zeros of (3+2p)(2p-3)(p+1)=0 are p=-3/2 ,p= 3/2 ,p=-1

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(3 + 2 p) (2 p - 3) (p + 1) = 0

p = - \frac{3}{2}, p = \frac{3}{2}, p = - 1

Decimal Notation

p = - 1.5, p = 1.5, p = - 1

Solution steps

(3 + 2 p) (2 p - 3) (p + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 + 2 p = 0 or 2 p - 3 = 0 or p + 1 = 0

Solve

3 + 2 p = 0 : p = - \frac{3}{2}

Solve

2 p - 3 = 0 : p = \frac{3}{2}

Solve

p + 1 = 0 : p = - 1

The solutions are

p = - \frac{3}{2}, p = \frac{3}{2}, p = - 1

- 4 x^{4} - 16 x^{2} = 0

(1 + \frac{x}{12})^{5} = 243

x^{30} = 36

x^{8} + 3 x^{4} + 4 = 0

x, x^{3} - 1 = 0

What are the solutions to the equation (3+2p)(2p-3)(p+1)=0 ?
The solutions to the equation (3+2p)(2p-3)(p+1)=0 are p=-3/2 ,p= 3/2 ,p=-1
Find the zeros of (3+2p)(2p-3)(p+1)=0
The zeros of (3+2p)(2p-3)(p+1)=0 are p=-3/2 ,p= 3/2 ,p=-1