and remainder is 0.
![\frac{5x^4+5x^2+5}{x^2-x+1}\\=\frac{5(x^4+x^2+1)}{x^2-x+1}\\=\frac{5[x^2(x^2+1)+1]}{x^2-x+1}\\=\frac{5[x^2(x^2-x+x+1)+1]}{x^2-x+1}\\=\frac{5[x^2(x^2-x+1)+x^3+1]}{x^2-x+1}\\=\frac{5[x^2(x^2-x+1)+x(x^2)+1]}{x^2-x+1}\\=\frac{5[x^2(x^2-x+1)+x(x^2-x+1+x-1)+1]}{x^2-x+1}\\=\frac{5[x^2(x^2-x+1)+x(x^2-x+1)+(x^2-x+1)]}{x^2-x+1}\\=\frac{5[(x^2-x+1)(x^2+x+1)}{x^2-x+1} \\=5(x^2+x+1)](/tpl/images/0655/4113/38cf9.png)
What
are the quotient and remainder of (5x^4+5x^2 +5)/(x^2-x+1)?...
Mathematics, 19.05.2020 15:20 1tzM3
What
are the quotient and remainder of (5x^4+5x^2 +5)/(x^2-x+1)?
Answers: 3
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