Question

rewrite

\\\frac{-20x^4 + 5x^3 + 20x^2 + 3x - 3}{4x^2 - x - 4}\\

in the form

\\q(x) + \frac{r(x)}{4x^2 - x - 4}.\\

\\\frac{-20x^4 + 5x^3 + 20x^2 + 3x - 3}{4x^2 - x - 4} = \text{ } + \frac{\text{ }}{4x^2 - x - 4}\\

Explanation:

⚡ Using what you learned: dividing polynomials

Step 1: Set up polynomial long division

Divide the numerator by the denominator:

$$ (-20x^4 + 5x^3 + 20x^2 + 3x - 3) \div (4x^2 - x - 4) $$

Step 2: Divide the first term

Divide the leading term of the dividend by the leading term of the divisor:

$$ \frac{-20x^4}{4x^2} = -5x^2 $$

Multiply \(-5x^2\) by the divisor:

$$ -5x^2(4x^2 - x - 4) = -20x^4 + 5x^3 + 20x^2 $$

Subtract this from the dividend:

$$ (-20x^4 + 5x^3 + 20x^2 + 3x - 3) - (-20x^4 + 5x^3 + 20x^2) = 3x - 3 $$

Step 3: Identify quotient and remainder

Since the degree of the remaining polynomial \(3x - 3\) is less than the degree of the divisor \(4x^2 - x - 4\), the division is complete.

• Quotient \(q(x) = -5x^2\)
• Remainder \(r(x) = 3x - 3\)

Answer:

$$ \frac{-20x^4 + 5x^3 + 20x^2 + 3x - 3}{4x^2 - x - 4} = -5x^2 + \frac{3x - 3}{4x^2 - x - 4} $$