Question

divide the polynomial and simplify:
x + 1 \overline{) x^2 - 9x - 10}

x^2 + 3x - 5 \overline{) x^4 + 4x^3 + x - 10}

Explanation:

Step1: Divide first polynomial pair

We use polynomial long division for $\frac{x^2 -9x -10}{x+1}$. First, divide $x^2$ by $x$:
$\frac{x^2}{x}=x$
Multiply $x+1$ by $x$: $x(x+1)=x^2+x$
Subtract from dividend: $(x^2-9x-10)-(x^2+x)=-10x-10$
Divide $-10x$ by $x$: $\frac{-10x}{x}=-10$
Multiply $x+1$ by $-10$: $-10(x+1)=-10x-10$
Subtract: $(-10x-10)-(-10x-10)=0$

Step2: Divide second polynomial pair

For $\frac{x^4 +4x^3 +x -10}{x^2+3x-5}$, divide $x^4$ by $x^2$:
$\frac{x^4}{x^2}=x^2$
Multiply $x^2+3x-5$ by $x^2$: $x^2(x^2+3x-5)=x^4+3x^3-5x^2$
Subtract from dividend: $(x^4+4x^3+x-10)-(x^4+3x^3-5x^2)=x^3+5x^2+x-10$
Divide $x^3$ by $x^2$: $\frac{x^3}{x^2}=x$
Multiply $x^2+3x-5$ by $x$: $x(x^2+3x-5)=x^3+3x^2-5x$
Subtract: $(x^3+5x^2+x-10)-(x^3+3x^2-5x)=2x^2+6x-10$
Divide $2x^2$ by $x^2$: $\frac{2x^2}{x^2}=2$
Multiply $x^2+3x-5$ by $2$: $2(x^2+3x-5)=2x^2+6x-10$
Subtract: $(2x^2+6x-10)-(2x^2+6x-10)=0$

Answer:

1. $\frac{x^2 -9x -10}{x+1}=x-10$
2. $\frac{x^4 +4x^3 +x -10}{x^2+3x-5}=x^2+x+2$