Question

What
are the quotient and remainder of (5x^4+5x^2 +5)/(x^2-x+1)?

Answer 1

Quotient is
`5(x^2+x+1)` and remainder is 0.

Explanation:

Given:
`(5x^4+5x^2+5)/(x^2-x+1)`

To find: quotient and remainder

Solution:

In the given question,

Dividend =
`5x^4+5x^2+5`

Divisor =
`x^2-x+1`

`(5x^4+5x^2+5)/(x^2-x+1)\\=(5(x^4+x^2+1))/(x^2-x+1)\\=(5[x^2(x^2+1)+1])/(x^2-x+1)\\=(5[x^2(x^2-x+x+1)+1])/(x^2-x+1)\\=(5[x^2(x^2-x+1)+x^3+1])/(x^2-x+1)\\=(5[x^2(x^2-x+1)+x(x^2)+1])/(x^2-x+1)\\=(5[x^2(x^2-x+1)+x(x^2-x+1+x-1)+1])/(x^2-x+1)\\=(5[x^2(x^2-x+1)+x(x^2-x+1)+(x^2-x+1)])/(x^2-x+1)\\=(5[(x^2-x+1)(x^2+x+1))/(x^2-x+1) \\=5(x^2+x+1)`

So, quotient is
`5(x^2+x+1)` and remainder is 0.