What are the solutions to the equation (x-1)(x+2)-x^2(x+2)=21 ?

The solutions to the equation (x-1)(x+2)-x^2(x+2)=21 are x\approx-3.34911…

Find the zeros of (x-1)(x+2)-x^2(x+2)=21

The zeros of (x-1)(x+2)-x^2(x+2)=21 are x\approx-3.34911…

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(x - 1) (x + 2) - x^{2} (x + 2) = 21

x \approx - 3.34911 \dots

Solution steps

(x - 1) (x + 2) - x^{2} (x + 2) = 21

Expand

(x - 1) (x + 2) - x^{2} (x + 2) : - x^{3} - x^{2} + x - 2

- x^{3} - x^{2} + x - 2 = 21

Move

21

to the left side

- x^{3} - x^{2} + x - 23 = 0

Find one solution for

- x^{3} - x^{2} + x - 23 = 0

using Newton-Raphson:

x \approx - 3.34911 \dots

Apply long division:

\frac{- x ^{3} - x ^{2} + x - 23}{x + 3.34911 \dots} = - x^{2} + 2.34911 \dots x - 6.86747 \dots

- x^{2} + 2.34911 \dots x - 6.86747 \dots \approx 0

Find one solution for

- x^{2} + 2.34911 \dots x - 6.86747 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx - 3.34911 \dots

e (x) = (- x + x^{2})^{2} + 4 x^{3}

3 x^{4} - 24 x^{3} + 48 x^{2} + x - 4 = 0

x, x^{4} + y^{4} = 0

x^{3} - 5 x^{2} + 8 x - 4 = 0

x^{3} + 2 x^{2} - 5 x - 6 = 0

What are the solutions to the equation (x-1)(x+2)-x^2(x+2)=21 ?
The solutions to the equation (x-1)(x+2)-x^2(x+2)=21 are x\approx-3.34911…
Find the zeros of (x-1)(x+2)-x^2(x+2)=21
The zeros of (x-1)(x+2)-x^2(x+2)=21 are x\approx-3.34911…