What are the solutions to the equation (v-1)(v^2+v-4)=0 ?

The solutions to the equation (v-1)(v^2+v-4)=0 are v=1,v=(-1+sqrt(17))/2 ,v=(-1-sqrt(17))/2

Find the zeros of (v-1)(v^2+v-4)=0

The zeros of (v-1)(v^2+v-4)=0 are v=1,v=(-1+sqrt(17))/2 ,v=(-1-sqrt(17))/2

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(v - 1) (v^{2} + v - 4) = 0

v = 1, v = \frac{- 1 + 17}{2}, v = \frac{- 1 - 17}{2}

Solution steps

(v - 1) (v^{2} + v - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v - 1 = 0 or v^{2} + v - 4 = 0

Solve

v - 1 = 0 : v = 1

Solve

v^{2} + v - 4 = 0 : v = \frac{- 1 + 17}{2}, v = \frac{- 1 - 17}{2}

The solutions are

v = 1, v = \frac{- 1 + 17}{2}, v = \frac{- 1 - 17}{2}

4 x^{4} + 2 x^{2} - 56 = 0

0 = 2 x^{3} - 3 x + 1

0 = (x - 1)^{3} - 2

x^{3} - 15 x = 0

- 3 x^{2} x = - 7

What are the solutions to the equation (v-1)(v^2+v-4)=0 ?
The solutions to the equation (v-1)(v^2+v-4)=0 are v=1,v=(-1+sqrt(17))/2 ,v=(-1-sqrt(17))/2
Find the zeros of (v-1)(v^2+v-4)=0
The zeros of (v-1)(v^2+v-4)=0 are v=1,v=(-1+sqrt(17))/2 ,v=(-1-sqrt(17))/2