What are the solutions to the equation (x+1)(x+2)(x+3)(x+4)-8=0 ?

The solutions to the equation (x+1)(x+2)(x+3)(x+4)-8=0 are x=-(-sqrt(17)+5)/2 ,x=-(sqrt(17)+5)/2

Find the zeros of (x+1)(x+2)(x+3)(x+4)-8=0

The zeros of (x+1)(x+2)(x+3)(x+4)-8=0 are x=-(-sqrt(17)+5)/2 ,x=-(sqrt(17)+5)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0

x = - \frac{- 17 + 5}{2}, x = - \frac{17 + 5}{2}

Decimal Notation

x = - 0.43844 \dots, x = - 4.56155 \dots

Solution steps

(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0

Expand

(x + 1) (x + 2) (x + 3) (x + 4) - 8 : x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 16

x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 16 = 0

Find one solution for

x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 16 = 0

using Newton-Raphson:

x \approx - 0.43844 \dots

Apply long division:

\frac{x ^{4} + 10 x ^{3} + 35 x ^{2} + 50 x + 16}{x + 0.43844 \dots} = x^{3} + 9.56155 \dots x^{2} + 30.80776 \dots x + 36.49242 \dots

x^{3} + 9.56155 \dots x^{2} + 30.80776 \dots x + 36.49242 \dots \approx 0

Find one solution for

x^{3} + 9.56155 \dots x^{2} + 30.80776 \dots x + 36.49242 \dots = 0

using Newton-Raphson:

x \approx - 4.56155 \dots

Apply long division:

\frac{x ^{3} + 9.56155 \dots x ^{2} + 30.80776 \dots x + 36.49242 \dots}{x + 4.56155 \dots} = x^{2} + 5 x + 8

x^{2} + 5 x + 8 \approx 0

Find one solution for

x^{2} + 5 x + 8 = 0

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx - 0.43844 \dots : x = - \frac{- 17 + 5}{2}

x \approx - 4.56155 \dots : x = - \frac{17 + 5}{2}

x = - \frac{- 17 + 5}{2}, x = - \frac{17 + 5}{2}

The solutions are

x = - \frac{- 17 + 5}{2}, x = - \frac{17 + 5}{2}

3 x^{3} = 375

x^{3} - 8 x^{2} + 15 x = 0

(1 - x)^{4} + (1 + x)^{4} = 82

x^{4} + 5 x^{2} + 6 = 0

m^{3} + 3 m^{2} - 2 = 3 m^{3}

What are the solutions to the equation (x+1)(x+2)(x+3)(x+4)-8=0 ?
The solutions to the equation (x+1)(x+2)(x+3)(x+4)-8=0 are x=-(-sqrt(17)+5)/2 ,x=-(sqrt(17)+5)/2
Find the zeros of (x+1)(x+2)(x+3)(x+4)-8=0
The zeros of (x+1)(x+2)(x+3)(x+4)-8=0 are x=-(-sqrt(17)+5)/2 ,x=-(sqrt(17)+5)/2