Answer:
Quotient is [tex]5(x^2+x+1)[/tex] and remainder is 0.
Step-by-step explanation:
Given: [tex]\frac{5x^4+5x^2+5}{x^2-x+1}[/tex]
To find: quotient and remainder
Solution:
In the given question,
Dividend = [tex]5x^4+5x^2+5[/tex]
Divisor = [tex]x^2-x+1[/tex]
[tex]\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)[/tex]
So, quotient is [tex]5(x^2+x+1)[/tex] and remainder is 0.
