Question

factories the following​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

• x⁴ + x² + 1

It is certainly possible over the real numbers, because in this setting the only nonlinear irreducible polynomials have degree two (and negative discriminant).

Now, It's time to Factories this Polynomial.

We can add & subtract x² to make Factorisation easier as that:-

$\\ \circ \ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \colon\Rightarrow{\sf{ x^4+x^2+1}} \\ \\ \colon\Rightarrow{\sf{ (x^4+x^2+1+x^2)-x^2 }} \\ \\ \colon\Rightarrow{\sf{ (x^4+2x^2+1)-x^2 }} \\ \\ \colon\Rightarrow{\sf{ ([x^2]^2+2[x^2][1]+[1]^2)-x^2 }} \\ \\ \colon\Rightarrow{\sf{ (x^2+1)^2-x^2 }}$

Now, Using this Identity as:

$\colon\Rightarrow{\sf\purple{ a^2-b^2 = (a-b)(a+b) }} \\ \\ \colon\Rightarrow{\sf{ (x^2+1)^2-x^2 }} \\ \\ \colon\Rightarrow{\sf{ (x^2+1-x)(x^2+1+x) }} \\ \\ \colon\Rightarrow{\underline{\boxed{\sf{ (x^2-x+1)(x^2+x+1) }}}}$

Hence,

$\\ {\pmb{\underline{\sf\red{ (x^4+x^2+1) = (x^2-x+1)(x^2+x+1) \ after \ Arranging. }}}}$