Respuesta :
Answer:
[tex]\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}=x^3-3x^2+7x-22+\frac{70}{x+3}[/tex]
Step-by-step explanation:
We are going to use long division to find the expression for [tex]\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}[/tex]
Step 1: Divide [tex]\frac{x^4-2x^2-x+4}{x+3}[/tex]
- Divide the leading coefficients of the numerator [tex]x^4-2x^2-x+4[/tex] and the divisor [tex]x+3[/tex]
[tex]\frac{x^4}{x}=x^3[/tex]
Quotient = [tex]x^3[/tex]
- Multiply [tex]x+3[/tex] by [tex]x^3[/tex]
[tex]x^4+3x^3[/tex]
- Subtract [tex]x^4+3x^3[/tex] from [tex]x^4-2x^2-x+4[/tex] to get new remainder
Remainder = [tex]-3x^3-2x^2-x+4[/tex]
Therefore,
[tex]\frac{x^4-2x^2-x+4}{x+3}=x^3+\frac{-3x^3-2x^2-x+4}{x+3}[/tex]
Step 2: Divide [tex]\frac{-3x^3-2x^2-x+4}{x+3}[/tex]
Quotient = [tex]-3x^2[/tex]
Remainder = [tex]7x^2-x+4[/tex]
Therefore,
[tex]\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+\frac{7x^2-x+4}{x+3}[/tex]
Step 3: Divide [tex]\frac{7x^2-x+4}{x+3}[/tex]
Quotient = [tex]7x[/tex]
Remainder = [tex]-22x+4[/tex]
Therefore,
[tex]\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x+\frac{-22x+4}{x+3}[/tex]
Step 4:
Quotient = [tex]-22[/tex]
Remainder = [tex]70[/tex]
Therefore,
[tex]\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x-22+\frac{70}{x+3}[/tex]
