x-2-x-1-x-5-8-x-1-

Question Number 176458 by mathlove last updated on 19/Sep/22

x^{2} + x = 1

\frac{x^{5} + 8}{x + 1} = ?

Answered by Rasheed.Sindhi last updated on 19/Sep/22

x^{2} + x = 1; \frac{x^{5} + 8}{x + 1} = ?

x^{2} = 1 - x \Rightarrow x^{3} = x - x^{2} = x - (1 - x) = 2 x - 1

\Rightarrow x^{5} = 2 x^{3} - x^{2} = 2 (2 x - 1) - (1 - x)

\Rightarrow x^{5} = 5 x - 3

\frac{x^{5} + 8}{x + 1} = \frac{5 x - 3 + 8}{x + 1} = \frac{5 x + 5}{x + 1} = 5 (\frac{x + 1}{x + 1}) = 5

Commented by mathlove last updated on 20/Sep/22

t h a n k s

Answered by Rasheed.Sindhi last updated on 20/Sep/22

\underset{(R e p l a c i n g 1^{'} \boldsymbol s)}{\boldsymbol AnOther \boldsymbol way \dots}

\begin{matrix} x^{2} + x = 1; \frac{x^{5} + 8}{x + 1} = ? \end{matrix}

\frac{x^{5} + 8}{x + 1} = \frac{x^{5} + 8 (x^{2} + x)}{x + x^{2} + x} = \frac{x^{5} + 8 x^{2} + 8 x}{x^{2} + 2 x}

= \frac{x^{4} + 8 x + 8}{x + 2} = \frac{x^{4} + 8 x + 8 (x^{2} + x)}{x + 2 (x^{2} + x)} = \frac{x^{4} + 8 x^{2} + 16 x}{2 x^{2} + 3 x}

= \frac{x^{3} + 8 x + 16}{2 x + 3} = \frac{x^{3} + 8 x + 16 (x^{2} + x)}{2 x + 3 (x^{2} + x)} = \frac{x^{3} + 16 x^{2} + 24 x}{3 x^{2} + 5 x}

= \frac{x^{2} + 16 x + 24}{3 x + 5} = \frac{x^{2} + 16 x + 24 (x^{2} + x)}{3 x + 5 (x^{2} + x)}

= \frac{25 x^{2} + 40 x}{5 x^{2} + 8 x} = \frac{25 x + 40}{5 x + 8} = \frac{5 \cancel (5 x + 8)}{\cancel (5 x + 8)}

= 5

Answered by Rasheed.Sindhi last updated on 20/Sep/22

Prime causes double exponent: use braces to clarify

x^{2} + x = 1; \frac{x^{5} + 8}{x + 1} = ?

x^{2} = 1 - x

\frac{x^{5} + 8}{x + 1} = \frac{x {(x^{2})}^{2} + 8}{x + 1} = \frac{x {(1 - x)}^{2} + 8}{x + 1}

= \frac{x - 2 x^{2} + x^{3} + 8}{x + 1} = \frac{x - x^{2} (2 - x) + 8}{x + 1}

= \frac{x - (1 - x) (2 - x) + 8}{x + 1}

= \frac{x - 2 + x + 2 x - x^{2} + 8}{x + 1} = \frac{- x^{2} + 4 x + 6}{x + 1}

= \frac{- (1 - x) + 4 x + 6}{x + 1} = \frac{- 1 + x + 4 x + 6}{x + 1}

= \frac{5 x + 5}{x + 1} = \frac{5 \cancel (x + 1)}{\cancel (x + 1)} = 5

Answered by Rasheed.Sindhi last updated on 20/Sep/22

A_{\boldsymbol pproach}^{\boldsymbol nOther} \dots

x^{2} + x = 1; \frac{x^{5} + 8}{x + 1} = ?

\begin{matrix} x (x + 1) = 1 \end{matrix}

\frac{x^{5} + 8}{x + 1}

= \frac{x (x^{5} + 8)}{x (x + 1)} [x (x + 1) = 1]

= x^{6} + x^{5} - x^{5} + 8 x

= x^{5} (x + 1) - x^{5} + 8 x

= x^{4} \cdot x (x + 1) - x^{5} + 8 x

= x^{4} + x^{4} - x^{5} - x^{4} + 8 x

= 2 x^{4} - x^{3} . x (x + 1) + 8 x

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

= 2 x^{2} . x (x + 1) - 3 x^{3} + 8 x

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

= 5 x^{2} - 3 x . x (x + 1) + 8 x

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

= 5 x^{2} + 5 x

= 5 x (x + 1)

= 5

Commented by Tawa11 last updated on 20/Sep/22

Great sir

Commented by Rasheed.Sindhi last updated on 20/Sep/22

T \boldsymbol han k \boldsymbol s \boldsymbol miss

Commented by mathlove last updated on 02/Oct/22

t h a n k s a l o t

Answered by Peace last updated on 20/Sep/22

x^{2} = 1 - x, x^{3} = x - x^{2} = 2 x - 1

x^{4} = 2 x^{2} - x = 2 - 3 x

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

\frac{x^{5} + 8}{x + 1} = \frac{5 x - 3 + 8}{x + 1} = 5

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