What are the solutions to the equation (Q^3)/3-10Q^2+80Q+648=4Q ?

The solutions to the equation (Q^3)/3-10Q^2+80Q+648=4Q are Q\approx-4.88120…

Find the zeros of (Q^3)/3-10Q^2+80Q+648=4Q

The zeros of (Q^3)/3-10Q^2+80Q+648=4Q are Q\approx-4.88120…

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\frac{Q ^{3}}{3} - 10 Q^{2} + 80 Q + 648 = 4 Q

Q \approx - 4.88120 \dots

Solution steps

\frac{Q ^{3}}{3} - 10 Q^{2} + 80 Q + 648 = 4 Q

Multiply both sides by

3

Q^{3} - 30 Q^{2} + 240 Q + 1944 = 12 Q

Move

12 Q

to the left side

Q^{3} - 30 Q^{2} + 228 Q + 1944 = 0

Find one solution for

Q^{3} - 30 Q^{2} + 228 Q + 1944 = 0

using Newton-Raphson:

Q \approx - 4.88120 \dots

Apply long division:

\frac{Q ^{3} - 30 Q ^{2} + 228 Q + 1944}{Q + 4.88120 \dots} = Q^{2} - 34.88120 \dots Q + 398.26231 \dots

Q^{2} - 34.88120 \dots Q + 398.26231 \dots \approx 0

Find one solution for

Q^{2} - 34.88120 \dots Q + 398.26231 \dots = 0

using Newton-Raphson:

No Solution for

Q \in R

The solution is

Q \approx - 4.88120 \dots

x^{3} + x^{2} - 17 x + 15

x^{3} - x^{2} - 25 x + 25 = 0

- 19 x^{3} - 12 x^{2} + 16 x = 0

3 y^{3} - 20 y^{2} - 7 y = 0

4 t^{3} - 12 t^{2} - 72 t + 25 = 25

What are the solutions to the equation (Q^3)/3-10Q^2+80Q+648=4Q ?
The solutions to the equation (Q^3)/3-10Q^2+80Q+648=4Q are Q\approx-4.88120…
Find the zeros of (Q^3)/3-10Q^2+80Q+648=4Q
The zeros of (Q^3)/3-10Q^2+80Q+648=4Q are Q\approx-4.88120…