Question

Decompose the polynomial x^3-2x^2-5x+6 into linear factors, given that x-1 is a factor

Answer 1

To decompose the polynomial \( x^3 - 2x^2 - 5x + 6 \) into linear factors, given that \( x - 1 \) is a factor, we can use polynomial long division to divide the cubic polynomial by \( x - 1 \).

### Step 1: Polynomial Long Division
We divide \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \).

1. Divide the leading term: \( \frac{x^3}{x} = x^2 \).
2. Multiply \( x^2 \) by \( x - 1 \):
\[
x^2 \cdot (x - 1) = x^3 - x^2.
\]
3. Subtract:
\[
\begin{align*}
(x^3 - 2x^2) - (x^3 - x^2) &= -2x^2 + x^2 = -x^2.
\end{align*}
\]
4. Bring down the next term:
\[
-x^2 - 5x \Rightarrow -x^2 - 5x.
\]
5. Divide the leading term: \( \frac{-x^2}{x} = -x \).
6. Multiply \( -x \) by \( x - 1 \):
\[
-x \cdot (x - 1) = -x^2 + x.
\]
7. Subtract:
\[
(-x^2 - 5x) - (-x^2 + x) = -5x - x = -6x.
\]
8. Bring down the next term:
\[
-6x + 6.
\]
9. Divide the leading term: \( \frac{-6x}{x} = -6 \).
10. Multiply \( -6 \) by \( x - 1 \):
\[
-6 \cdot (x - 1) = -6x + 6.
\]
11. Subtract:
\[
(-6x + 6) - (-6x + 6) = 0.
\]

The division is complete, and we find that:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6).
\]

### Step 2: Factor the Quadratic
Now we need to factor \( x^2 - x - 6 \).

To factor \( x^2 - x - 6 \), we can look for two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of \(x\)). These numbers are \(-3\) and \(2\).

So we can factor it as:
\[
x^2 - x - 6 = (x - 3)(x + 2).
\]

### Step 3: Write the Fully Factored Form
Combining the factors, we find:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2).
\]

### Final Answer
The polynomial \( x^3 - 2x^2 - 5x + 6 \) is decomposed into linear factors as:
\[
(x - 1)(x - 3)(x + 2).
\]