Question

Find the quotient. Justify your answer. x^5+2x^4-7x^2-19x+15/x^2+2x+5

Answer 1

Answer:

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3$

Step-by-step explanation:

To find the quotient we need to apply long division on $\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}$, as follows:

1. Divide $\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $x^5+2x^4-7x^2-19x+15$ and the divisor $x^2+2x+5$

$\frac{x^5}{x^2}=x^3$

Quotient = $x^3$

Multiply $x^2+2x+5$ by $x^3$ = $x^5+2x^4+5x^3$

Subtract $x^5+2x^4+5x^3$ from $x^5+2x^4-7x^2-19x+15$ to get new remainder

Remainder = $-5x^3-7x^2-19x+15$

Therefore

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3+\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}$

      2. Divide $\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $-5x^3-7x^2-19x+15$ and the divisor $x^2+2x+5$

$\frac{-5x^3}{x^2}=-5x$

Quotient = $-5x$

Multiply $x^2+2x+5$ by $-5x$ = $-5x^3-10x^2-25x$

Subtract $-5x^3-10x^2-25x$ from $-5x^3-7x^2-19x+15$ to get new remainder

Remainder = $3x^2+6x+15$

Therefore

$\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}=-5x+\frac{3x^2+6x+15}{x^2+2x+5}$

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3-5x+\frac{3x^2+6x+15}{x^2+2x+5}$

     3. Divide $\frac{3x^2+6x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $3x^2+6x+15$ and the divisor $x^2+2x+5$

$\frac{3x^2}{x^2}=3$

Quotient = 3

Multiply $x^2+2x+5$ by 3 = $\:3x^2+6x+15$

Subtract $\:3x^2+6x+15$ from $3x^2+6x+15$ to get new remainder

Remainder = 0

Therefore

$\frac{3x^2+6x+15}{x^2+2x+5}=3$

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3$