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

Answer:[tex]\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3[/tex]Step-by-step explanation:To find the quotient we need to apply long division on [tex]\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}[/tex], as follows:Divide [tex]\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}[/tex]Divide the leading coefficients of the numerator [tex]x^5+2x^4-7x^2-19x+15[/tex] and the divisor [tex]x^2+2x+5[/tex][tex]\frac{x^5}{x^2}=x^3[/tex]Quotient = [tex]x^3[/tex]Multiply [tex]x^2+2x+5[/tex] by [tex]x^3[/tex] = [tex]x^5+2x^4+5x^3[/tex]Subtract [tex]x^5+2x^4+5x^3[/tex] from [tex]x^5+2x^4-7x^2-19x+15[/tex] to get new remainderRemainder = [tex]-5x^3-7x^2-19x+15[/tex]Therefore [tex]\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}[/tex]       2. Divide [tex]\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}[/tex]Divide the leading coefficients of the numerator [tex]-5x^3-7x^2-19x+15[/tex] and the divisor [tex]x^2+2x+5[/tex][tex]\frac{-5x^3}{x^2}=-5x[/tex]Quotient = [tex]-5x[/tex]Multiply [tex]x^2+2x+5[/tex] by [tex]-5x[/tex] = [tex]-5x^3-10x^2-25x[/tex]Subtract [tex]-5x^3-10x^2-25x[/tex] from [tex]-5x^3-7x^2-19x+15[/tex] to get new remainderRemainder = [tex]3x^2+6x+15[/tex]Therefore[tex]\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}=-5x+\frac{3x^2+6x+15}{x^2+2x+5}[/tex][tex]\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}[/tex]      3. Divide [tex]\frac{3x^2+6x+15}{x^2+2x+5}[/tex]Divide the leading coefficients of the numerator [tex]3x^2+6x+15[/tex] and the divisor [tex]x^2+2x+5[/tex][tex]\frac{3x^2}{x^2}=3[/tex]Quotient = 3Multiply [tex]x^2+2x+5[/tex] by 3 = [tex]\:3x^2+6x+15[/tex]Subtract [tex]\:3x^2+6x+15[/tex] from [tex]3x^2+6x+15[/tex] to get new remainderRemainder = 0Therefore [tex]\frac{3x^2+6x+15}{x^2+2x+5}=3[/tex][tex]\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3[/tex]