Question

P(x)=xcube plus xsquare plus x pulse 1 ÷ xsquare minus 1

Answer 1

Answer:

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

Step-by-step explanation:

$Given \:p(x)=\frac{x^{3}+x^{2}+x+1}{x^{2}-1}$

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

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

$=\frac{x^{2}+1}{x-1}$

Therefore,

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

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Answer 2

Solution

according to given question:-

P(x)=xcube plus xsquare plus x pulse 1 ÷ xsquare minus 1

Hence

p (x) = x³ + x² + x + 1 / x² - 1

= x² (x + 1) + 1 (x + 1) / (x² - 1 )²

= (x + 1) (x² + 1) / (x + 1) (x - 21)

Using Algebric Identity

= x² + 1 / x - 1

Hence,

p(x) = 2³ + x² + x + 1 / x² - 1

= x² + 1 / x - 1