# How do you divide (x^4-5x^3+6x^2-10x+15)/(x^2-7)?

Jun 3, 2018

#### Explanation:

Remember to write any terms whose coefficient is 0:
color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) color(white)( (x^4-5x^3+6x^2-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)

Because ${x}^{2} \cdot {x}^{2} = {x}^{4}$, the first term of the quotient is ${x}^{2}$:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2color(white)( 5x^3+6x^2-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)

Multiply the first term in the quotient by the divisor and subtract underneath:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2color(white)( 5x^3+6x^2-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- {x}^{4} - 0 {x}^{3} + 7 {x}^{2}} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 5 {x}^{3} + 13 {x}^{2} - 10 x$

Because $- 5 x \cdot {x}^{2} = - 5 {x}^{3}$. the next term in the quotient is $- 5 x$:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2-5xcolor(white)(6x^2-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- {x}^{4} - 0 {x}^{3} + 7 {x}^{2}} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 5 {x}^{3} + 13 {x}^{2} - 10 x$

Multiply the second term in the quotient by the divisor and subtract underneath:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2-5xcolor(white)(6x^2-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- {x}^{4} - 0 {x}^{3} + 7 {x}^{2}} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 5 {x}^{3} + 13 {x}^{2} - 10 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{5 {x}^{3} + \textcolor{w h i t e}{.} 0 {x}^{2} - 45 x} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 13 {x}^{2} - 45 x + 15$

Because $13 \cdot {x}^{2} = 13 {x}^{2}$. the next term in the quotient is $13$:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2-5x+13color(white)(-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- {x}^{4} - 0 {x}^{3} + 7 {x}^{2}} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 5 {x}^{3} + 13 {x}^{2} - 10 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{5 {x}^{3} + \textcolor{w h i t e}{.} 0 {x}^{2} - 45 x} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 13 {x}^{2} - 45 x + 15$

Multiply the third term in the quotient by the divisor and subtract underneath:

color(white)( (x^2+0x-7)/color(black)(x^2+0x-7)) (x^2-5x+13color(white)(-10x+15))/( ")" color(white)(")")x^4-5x^3+6x^2-10x+15)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- {x}^{4} - 0 {x}^{3} + 7 {x}^{2}} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} - 5 {x}^{3} + 13 {x}^{2} - 10 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{5 {x}^{3} + \textcolor{w h i t e}{.} 0 {x}^{2} - 35 x} \text{ } \downarrow$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 13 {x}^{2} - 45 x + 15$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- 13 {x}^{2} \textcolor{w h i t e}{.} - 0 x + 91}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 45 x + 106$

The quotient is ${x}^{2} - 5 x + 13$ with a remainder of $- 45 x + 10$