Question

Use polynomial long division to divide 6x^3-5x^2+13x-4 by 3x-1

Answer 1

To divide \(6x^3 - 5x^2 + 13x - 4\) by \(3x - 1\) using polynomial long division, follow these steps:

1. **Set up the long division**: Write \(6x^3 - 5x^2 + 13x - 4\) under the long division symbol and \(3x - 1\) outside.

2. **Divide the leading terms**: Divide the leading term of the dividend (\(6x^3\)) by the leading term of the divisor (\(3x\)):

\[
\frac{6x^3}{3x} = 2x^2
\]

3. **Multiply and subtract**: Multiply the entire divisor \(3x - 1\) by \(2x^2\) and subtract the result from the original polynomial:

\[
2x^2(3x - 1) = 6x^3 - 2x^2
\]

Now, perform the subtraction:

\[
(6x^3 - 5x^2 + 13x - 4) - (6x^3 - 2x^2) = -5x^2 + 2x^2 + 13x - 4 = -3x^2 + 13x - 4
\]

4. **Repeat the process**: Now divide the leading term of the new polynomial \(-3x^2\) by the leading term of the divisor \(3x\):

\[
\frac{-3x^2}{3x} = -x
\]

Multiply \(3x - 1\) by \(-x\):

\[
-x(3x - 1) = -3x^2 + x
\]

Now, subtract this from \(-3x^2 + 13x - 4\):

\[
(-3x^2 + 13x - 4) - (-3x^2 + x) = 13x - x - 4 = 12x - 4
\]

5. **Repeat again**: Now divide the leading term \(12x\) by the leading term \(3x\):

\[
\frac{12x}{3x} = 4
\]

Multiply \(3x - 1\) by \(4\):

\[
4(3x - 1) = 12x - 4
\]

Now, perform the subtraction:

\[
(12x - 4) - (12x - 4) = 0
\]

6. **Conclusion**: Since the remainder is \(0\), we can conclude that:

\[
\frac{6x^3 - 5x^2 + 13x - 4}{3x - 1} = 2x^2 - x + 4
\]

Thus, the result of the polynomial long division is:

\[
2x^2 - x + 4
\]