Question

What will be the quotient and remainder if $(6x^4+3x^2-9+5x+5x^3)÷(x^2-1)$?

• A. Quotient = $(6x^4+5x+9)$
  Remainder = $(10x^2+1)$
• B. Remainder= $(6x^2+5x+9)$
  Quotient= $\left(10x\right)$
• C. Quotient = $(6x^4+5x+9)$
  Remainder = $\left(10x\right)$
• D. Quotient = $(6x^4+5x+9)$
  Remainder = $(x+1)$

Answer 1

The correct option is C Quotient = $(6x^4+5x+9)$

Remainder = $\left(10x\right)$
Given: $(6x^4+3x^2-9+5x+5x^3)÷(x^2-1)$
$\begin{array}{c}6x^2+5x+9\\ x^2-16x^4+5x^3+3x^2+5x-9\\ \underset{–}{6x^4-6x^2-+}2\\ 5x^3+9x^2+5x-9\\ \underset{–}{5x^3-5x-+}\\ 9x^2+10x-9\\ \underset{–}{9x^2-9-+}\\ \underset{–}{10x}\end{array}$
Quotient = $(6x^2+5x+9)$
Remainder = $\left(10x\right)$