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

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

Find the zeros of 4(v-4)(v+1)=0

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

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4 (v - 4) (v + 1) = 0

v = 4, v = - 1

Solution steps

4 (v - 4) (v + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v - 4 = 0 or v + 1 = 0

Solve

v - 4 = 0 : v = 4

Solve

v + 1 = 0 : v = - 1

The solutions to the quadratic equation are:

v = 4, v = - 1

3 x^{2} - 24 x + 48

- 65 y + 19 < - 2 y + 41

\frac{\frac{6}{7}}{6}

\frac{- 4 + 6 i}{2 + 7 i}

- 18 > - 4 x - 6

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