Question

Use long division to divide 8x3−20x2+18x−11 by 2x2−3 .(1 point)

Answer 1

To divide \( 8x^3 - 20x^2 + 18x - 11 \) by \( 2x^2 - 3 \) using long division, follow these steps:

1. Set up the long division:

\[
\begin{array}{r|l}
2x^2 - 3 & 8x^3 - 20x^2 + 18x - 11 \\
\end{array}
\]

2. Divide the leading term of the dividend \( 8x^3 \) by the leading term of the divisor \( 2x^2 \) to get \( 4x \) (since \( 8x^3 \div 2x^2 = 4x \)).

3. Multiply \( 4x \) by the entire divisor \( 2x^2 - 3 \):

\[
4x(2x^2 - 3) = 8x^3 - 12x
\]

4. Subtract this result from the original polynomial:

\[
(8x^3 - 20x^2 + 18x - 11) - (8x^3 - 12x) = -20x^2 + 18x + 12x - 11 = -20x^2 + 30x - 11
\]

5. Now, repeat the process. Divide the leading term of the new polynomial \( -20x^2 \) by the leading term of the divisor \( 2x^2 \):

\[
-20x^2 \div 2x^2 = -10
\]

6. Multiply \( -10 \) by the divisor \( 2x^2 - 3 \):

\[
-10(2x^2 - 3) = -20x^2 + 30
\]

7. Subtract this from \( -20x^2 + 30x - 11 \):

\[
(-20x^2 + 30x - 11) - (-20x^2 + 30) = 30x - 11 - 30 = 30x - 41
\]

8. At this point, we have:

\[
\begin{array}{r|l}
2x^2 - 3 & 8x^3 - 20x^2 + 18x - 11 \\
& \underline{4x} \\
& -20x^2 + 30x - 11 \\
& \underline{-10} \\
& 30x - 41 \\
\end{array}
\]

9. Now we can't divide anymore because the degree of the remainder \( 30x - 41 \) is less than the degree of the divisor \( 2x^2 - 3 \).

So, the final result of the division is:

\[
\text{Quotient: } 4x - 10
\]
\[
\text{Remainder: } 30x - 41
\]

Therefore, the overall result can be written as:

\[
8x^3 - 20x^2 + 18x - 11 = (2x^2 - 3)(4x - 10) + (30x - 41)
\]

In conclusion, the result of dividing \( 8x^3 - 20x^2 + 18x - 11 \) by \( 2x^2 - 3 \) is:

\[
\text{Quotient: } 4x - 10, \quad \text{Remainder: } 30x - 41
\]