What are the solutions to the equation x^3(3x+5)=7(x+4) ?

The solutions to the equation x^3(3x+5)=7(x+4) are x\approx-2.12372…,x\approx 1.58819…

Find the zeros of x^3(3x+5)=7(x+4)

The zeros of x^3(3x+5)=7(x+4) are x\approx-2.12372…,x\approx 1.58819…

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x^{3} (3 x + 5) = 7 (x + 4)

x \approx - 2.12372 \dots, x \approx 1.58819 \dots

Solution steps

x^{3} (3 x + 5) = 7 (x + 4)

Expand

x^{3} (3 x + 5) : 3 x^{4} + 5 x^{3}

Expand

7 (x + 4) : 7 x + 28

3 x^{4} + 5 x^{3} = 7 x + 28

Move

28

to the left side

3 x^{4} + 5 x^{3} - 28 = 7 x

Move

7 x

to the left side

3 x^{4} + 5 x^{3} - 28 - 7 x = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

3 x^{4} + 5 x^{3} - 7 x - 28 = 0

Find one solution for

3 x^{4} + 5 x^{3} - 7 x - 28 = 0

using Newton-Raphson:

x \approx - 2.12372 \dots

Apply long division:

\frac{3 x ^{4} + 5 x ^{3} - 7 x - 28}{x + 2.12372 \dots} = 3 x^{3} - 1.37118 \dots x^{2} + 2.91202 \dots x - 13.18435 \dots

3 x^{3} - 1.37118 \dots x^{2} + 2.91202 \dots x - 13.18435 \dots \approx 0

Find one solution for

3 x^{3} - 1.37118 \dots x^{2} + 2.91202 \dots x - 13.18435 \dots = 0

using Newton-Raphson:

x \approx 1.58819 \dots

Apply long division:

\frac{3 x ^{3} - 1.37118 \dots x ^{2} + 2.91202 \dots x - 13.18435 \dots}{x - 1.58819 \dots} = 3 x^{2} + 3.39341 \dots x + 8.30144 \dots

3 x^{2} + 3.39341 \dots x + 8.30144 \dots \approx 0

Find one solution for

3 x^{2} + 3.39341 \dots x + 8.30144 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx - 2.12372 \dots, x \approx 1.58819 \dots

x, x^{2} (5 - 2 x) = 4

x^{5} = (2 - x)^{3}

5 x^{4} = 3, x^{2} = 8

2 x^{5} - 6 x^{3} + 8 x^{2} - 5 = 0

(x^{3} - 4 x^{2}) = (16 x - 64)

What are the solutions to the equation x^3(3x+5)=7(x+4) ?
The solutions to the equation x^3(3x+5)=7(x+4) are x\approx-2.12372…,x\approx 1.58819…
Find the zeros of x^3(3x+5)=7(x+4)
The zeros of x^3(3x+5)=7(x+4) are x\approx-2.12372…,x\approx 1.58819…