factor by grouping

[Greta]

Factor the polynomial completely: x[supdhfxmh1]3[/supdhfxmh1]+x[supdhfxmh1]2[/supdhfxmh1]+x+1

x[supdhfxmh1]3[/supdhfxmh1]+x[supdhfxmh1]2[/supdhfxmh1]+x+1=x[supdhfxmh1]2[/supdhfxmh1](x+1)+(x+1) (factor by grouping)
=(x[supdhfxmh1]2[/supdhfxmh1]+1)(x+1) Is this the answer?

 

[lookagain]

Factor the polynomial completely: x[sup:3quvwobv]3[/sup:3quvwobv]+x[sup:3quvwobv]2[/sup:3quvwobv]+x+1

x[sup:3quvwobv]3[/sup:3quvwobv]+x[sup:3quvwobv]2[/sup:3quvwobv]+x+1=x[sup:3quvwobv]2[/sup:3quvwobv](x+1)+(x+1) (factor by grouping)
=(x[sup:3quvwobv]2[/sup:3quvwobv]+1)(x+1) Is this the answer?

Space things out. If you are to factor it over integer coefficients, then yes.

I advise showing, not necessarily in Latex, but with more spacing:

\(\displaystyle x^3 + x^2 + x + 1 =\)

\(\displaystyle x^2(x \ + \ 1) \ + \ 1(x \ + \ 1) =\)

\(\displaystyle ( See \ how \ I \ included \ \ a \ "1" \ in \ front \ of \ the \ second \ binomial \ factor \ as \ a \ place \ holder?)\)

\(\displaystyle (x^2 + 1)(x + 1) \ \ or \ \ (x + 1)(x^2 + 1)\)