Question

using long division, what is the quotient of $3x^4 + 20x^3 + 14x^2 + 17x + 30$ and $x + 6$?

a. $2x^3 + 14x^2 + 17x + 30$

b. $3x^3 - 2x^2 + 2x - 5$

c. $3x^3 + 2x^2 + 2x + 5$

d. $3x^2 + 2x + 2$

Explanation:

Step1: Divide the leading terms

Divide \(3x^4\) by \(x\) to get \(3x^3\). Multiply \(x + 6\) by \(3x^3\) to get \(3x^4 + 18x^3\). Subtract this from the original polynomial: \((3x^4 + 20x^3 + 14x^2 + 17x + 30) - (3x^4 + 18x^3) = 2x^3 + 14x^2 + 17x + 30\).

Step2: Divide the new leading term

Divide \(2x^3\) by \(x\) to get \(2x^2\). Multiply \(x + 6\) by \(2x^2\) to get \(2x^3 + 12x^2\). Subtract: \((2x^3 + 14x^2 + 17x + 30) - (2x^3 + 12x^2) = 2x^2 + 17x + 30\).

Step3: Divide the new leading term

Divide \(2x^2\) by \(x\) to get \(2x\). Multiply \(x + 6\) by \(2x\) to get \(2x^2 + 12x\). Subtract: \((2x^2 + 17x + 30) - (2x^2 + 12x) = 5x + 30\).

Step4: Divide the new leading term

Divide \(5x\) by \(x\) to get \(5\). Multiply \(x + 6\) by \(5\) to get \(5x + 30\). Subtract: \((5x + 30) - (5x + 30) = 0\).
The quotient is \(3x^3 + 2x^2 + 2x + 5\).

Answer:

C. \(3x^3 + 2x^2 + 2x + 5\)