Questions: For the given expression, find the quotient and the remainder. Check your work by verifying that (Quotient)(Divisor) + Remainder = Dividend. 9x^5-9x^2+6x+9 divided by 3x^3-1 Quotient: Remainder:

Solution Steps

To find the quotient and remainder of the given polynomial division, we can use polynomial long division or synthetic division. In this case, we'll use polynomial long division. The process involves dividing the leading term of the dividend by the leading term of the divisor to find the first term of the quotient. Then, multiply the entire divisor by this term and subtract the result from the dividend. Repeat the process with the new polynomial until the degree of the remainder is less than the degree of the divisor. Finally, verify the result by checking that the product of the quotient and divisor plus the remainder equals the original dividend.

We start with the dividend \( 9x^5 - 9x^2 + 6x + 9 \) and the divisor \( 3x^3 - 1 \).

Using polynomial long division, we divide the leading term of the dividend \( 9x^5 \) by the leading term of the divisor \( 3x^3 \) to get the first term of the quotient: \[ \frac{9x^5}{3x^3} = 3x^2 \] Next, we multiply the entire divisor \( 3x^3 - 1 \) by \( 3x^2 \): \[ 3x^2(3x^3 - 1) = 9x^5 - 3x^2 \] We subtract this from the original dividend: \[ (9x^5 - 9x^2 + 6x + 9) - (9x^5 - 3x^2) = -6x^2 + 6x + 9 \]

Now, we repeat the process with the new polynomial \( -6x^2 + 6x + 9 \). Divide the leading term \( -6x^2 \) by \( 3x^3 \): \[ \frac{-6x^2}{3x^3} = 0 \quad \text{(since the degree of the remainder is less than the degree of the divisor)} \] Thus, we stop here, and the remainder is \( -6x^2 + 6x + 9 \).

Final Answer