Question

Factorise:- x³+x²+x+1

Answer 1

Answer:

Hi bro

Step-by-step explanation:

First factor by grouping:

x3+x2−x−1=(x3+x2)−(x+1)=x2(x+1)−1(x+1)=(x2−1)(x+1)

Then notice that x2−1=x2−12 is a difference of squares, so we can use the difference of squares identity [ a2−b2=(a−b)(a+b) ] to find:

(x2−1)(x+1)=(x−1)(x+1)(x+1)=(x−1)(x+1)2

Alternatively, notice that the sum of the coefficients (1+1−1−1) is 0, so x=1 is a zero of this cubic polynomial and (x−1) is a factor.

Divide x3+x2−x−1 by (x−1) to get x2

Answer 2

$\LARGE{\underline{\underline{\bf{Solution :}}}}$

$\rule{200}{2}$

$\large{\leadsto{\underline{\underline{\bf{ \: To \: Find:}}}}}$

As, we have to factorise the given polynomial.

So, we have to find the value of x.

$\rule{200}{2}$

$\large{\leadsto{\underline{\underline{\bf{Explanation :}}}}}$

$\sf{→x^3 + x^2 + x + 1 = 0} \\ \\ \bf{\gray{Taking \: x^2 \: common \: from \: first \: two \: terms}} \\ \\ \sf{→x^2(x + 1) + x + 1} \\ \\ \bf{\gray{Taking \: 1 \: common \: from \: last \: two \: terms}} \\ \\ \sf{→x^2(x + 1) + 1(x + 1)} \\ \\ \sf{→(x^2 + 1)(x + 1)} \\ \\ \sf{→x^2 + 1 = 0}\\ \\ \sf{→x^2 = -1} \\ \\ \sf{→x = \sqrt{-1}} \\ \\ \bf{ \: \: \: \:\: \: \: \:\: \: \: \:\: \: \: \:\: \: \: \:\: \:\: \: \: \:\: \: \: \:\: \: \: \:\: \: \: \:\: \:\: \: \: Or\: \: \: \:\: \: \: \:\: \: \: \:\: \: \: \:\: \: \: \:} \\ \\ \sf{→x + 1 = 0} \\ \\ \sf{→x = -1}$

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