Question

x^5-2x²-x+2 factorise
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Answer 1

Given: The equation is$x^{5} -2x^{2} -x+2$.

We have to factorize the equation$x^{5} -2x^{2} -x+2$.

We are solving in the following way:

We have,$x^{5} -2x^{2} -x+2$

The alternate form of the given equation will be,

$=>(x+1)(x-1)(x-1)\left(x^{2}+x+2\right)\\=>(x-1)^{2}(x+1)\left(x^{2}+x+2\right)\\=>x\left(x^{4}-2 x-1\right)+2\\=>x\left(x\left(x^{3}-2\right)-1\right)+2$

We get the real roots as

$=>\begin{array}{l}x=-1 \\x=1\end{array}$

Hence, the factors are$(x+1)(x-1)(x-1)\left(x^{2}+x+2\right)$. and roots are$-1 and 1$.

Answer 2

Answer:

$x^5-2x^2-x+2=(x-1)^2(x+1)(x^2+x+2)$

Step-by-step explanation:

Given equation is $x^5-2x^2-x+2$.

We have to factorize the given equation.

Factorization or factoring consists of writing a number or another mathematical object as a product of several factors.

Rewrite the given equation as

$x^5-x-2x^2+2$

Now take $x$ as common from the first two terms and $-2$ as common from the last two terms,

$=x(x^4-1)-2(x^2-1)$

Now we know that $a^2-b^2=(a-b)(a+b)$,

$=x(x^2-1)(x^2+1)-2(x^2-1)$

Taking $x^2-1$ as common,

$=(x^2-1)(x(x^2+1)-2)$

$=(x-1)(x+1)(x^3+x-2)$

Now write $x$ as $2x-x$ in the last term,

$=(x-1)(x+1)(x^3+2x-x-2)$

$=(x-1)(x+1)(x^3-x+2x-2)$

Now factorize the third term,

$=(x-1)(x+1)(x(x^2-1)+2(x-1))$

$=(x-1)(x+1)(x(x-1)(x+1)+2(x-1))$

$=(x-1)^2(x+1)(x(x+1)+2)$

$=(x-1)^2(x+1)(x^2+x+2)$

After factorization, the equation becomes

$=(x-1)^2(x+1)(x^2+x+2)$