Answer:
(x+1) is a factor with remainder 0.
Step-by-step explanation:
We divide (x+1) into the polynomial [tex]-3x^4+2x^3-x^2+6[/tex] through long division or synthetic. We choose long division and look for what will multiply with (x+1) to make the polynomial [tex]-3x^4+2x^3-x^2+6[/tex] .
[tex](x+1)(-3x^3)=-3x^4-3x^3[/tex]
We subtract this from the original [tex]-3x^4-(-3x^4)+2x^3-(-3x^3)-x^2+6[/tex].
This leaves [tex]5x^3-x^2+6[/tex]. We repeat the step above.
[tex](x+1)(5x^2)=5x^3+5x^2[/tex].
We subtract this from [tex]5x^3-(5x^3)-x^2-(5x^2)+6=-6x^2+6[/tex]. We repeat the step above.
[tex](x+1)(-6x)=-6x^2-6x[/tex].
We subtract this from [tex]-6x^2-(-6x^2)+0x-(-6x)+6=6x+6[/tex]. We repeat the step above.
[tex](x+1)(6)=-6x+1[/tex].
We subtract this from [tex]6x-(6x)+6-(6)=0[/tex]. There is no remainder. This means (x+1) is a factor.
