Question

Use the long division method to find the result when 6, x, cubed, minus, 11, x, squared, minus, 26, x, plus, 156x
3
−11x
2
−26x+15 is divided by 3, x, plus, 53x+5.

Answer 1

To perform long division on the polynomial \(6x^3 - 11x^2 - 26x + 156\) by \(3x + 5\), we will follow these steps:

1. **Set up the long division**: Write \(6x^3 - 11x^2 - 26x + 156\) under the long division symbol and \(3x + 5\) outside the division symbol.

2. **Divide the leading term**: Divide the leading term of the dividend \(6x^3\) by the leading term of the divisor \(3x\):
\[
\frac{6x^3}{3x} = 2x^2
\]
This is the first term of the quotient.

3. **Multiply and subtract**: Multiply \(2x^2\) by the entire divisor \(3x + 5\):
\[
2x^2(3x + 5) = 6x^3 + 10x^2
\]
Now, subtract this from the original polynomial:
\[
(6x^3 - 11x^2 - 26x + 156) - (6x^3 + 10x^2) = -21x^2 - 26x + 156
\]

4. **Repeat the process**: Now divide the leading term of the new polynomial \(-21x^2\) by the leading term of the divisor \(3x\):
\[
\frac{-21x^2}{3x} = -7x
\]
This is the second term of the quotient.

5. **Multiply and subtract again**: Multiply \(-7x\) by the entire divisor:
\[
-7x(3x + 5) = -21x^2 - 35x
\]
Now perform the subtraction:
\[
(-21x^2 - 26x + 156) - (-21x^2 - 35x) = 9x + 156
\]

6. **Repeat the process once more**: Divide the leading term \(9x\) by the leading term \(3x\):
\[
\frac{9x}{3x} = 3
\]
This is the third term of the quotient.

7. **Multiply and subtract again**: Multiply \(3\) by the divisor:
\[
3(3x + 5) = 9x + 15
\]
Perform the subtraction:
\[
(9x + 156) - (9x + 15) = 141
\]

8. **Conclusion**: Since \(141\) (the remainder) has a lower degree than the divisor \(3x + 5\), we stop here.

The final result of the division is:

\[
2x^2 - 7x + 3 \quad \text{with a remainder of } 141.
\]

Thus, we can express the result as:
\[
\frac{6x^3 - 11x^2 - 26x + 156}{3x + 5} = 2x^2 - 7x + 3 + \frac{141}{3x + 5}.
\]