Question

If h(x)=x^4-2x^2-x+4 , then what is h(x)/(x+3)? You may use either long or synthetic division.

Answer 1

Answer:

$\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}=x^3-3x^2+7x-22+\frac{70}{x+3}$

Step-by-step explanation:

We are going to use long division to find the expression for $\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}$

Step 1: Divide $\frac{x^4-2x^2-x+4}{x+3}$

• Divide the leading coefficients of the numerator $x^4-2x^2-x+4$ and the divisor $x+3$

$\frac{x^4}{x}=x^3$

Quotient = $x^3$

• Multiply $x+3$ by $x^3$

$x^4+3x^3$

• Subtract $x^4+3x^3$ from $x^4-2x^2-x+4$ to get new remainder

Remainder = $-3x^3-2x^2-x+4$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3+\frac{-3x^3-2x^2-x+4}{x+3}$

Step 2: Divide $\frac{-3x^3-2x^2-x+4}{x+3}$

Quotient = $-3x^2$

Remainder = $7x^2-x+4$

Therefore,

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

Step 3: Divide $\frac{7x^2-x+4}{x+3}$

Quotient = $7x$

Remainder = $-22x+4$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x+\frac{-22x+4}{x+3}$

Step 4:

Quotient = $-22$

Remainder = $70$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x-22+\frac{70}{x+3}$