Question

Decompose the polynomial x3−2x2−5x+6 into its linear factors, given that x−1 is a facto

Answer 1

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

1. **Divide the polynomial \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \):**

**Step 1:** Divide the leading term of the polynomial by the leading term of the divisor:
\[
\frac{x^3}{x} = x^2
\]

**Step 2:** Multiply \( x^2 \) by \( x - 1 \):
\[
x^2(x - 1) = x^3 - x^2
\]

**Step 3:** Subtract this from the original polynomial:
\[
(x^3 - 2x^2 - 5x + 6) - (x^3 - x^2) = -2x^2 + x^2 - 5x + 6 = -x^2 - 5x + 6
\]

**Step 4:** Repeat the process with the new polynomial \(-x^2 - 5x + 6\):
Divide the leading term:
\[
\frac{-x^2}{x} = -x
\]

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

Subtract:
\[
(-x^2 - 5x + 6) - (-x^2 + x) = -5x - x + 6 = -6x + 6
\]

**Step 5:** Divide \(-6x + 6\):
\[
\frac{-6x}{x} = -6
\]

Multiply:
\[
-6(x - 1) = -6x + 6
\]

Subtract:
\[
(-6x + 6) - (-6x + 6) = 0
\]

2. **The result of the division** is:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6)
\]

3. **Now, factor \( x^2 - x - 6 \)**:
We need two numbers that multiply to \(-6\) and add to \(-1\). The pair that works is \( -3 \) and \( 2 \):
\[
x^2 - x - 6 = (x - 3)(x + 2)
\]

4. **Combine the factors**:
Therefore, the complete factorization of the polynomial is:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)
\]

Thus, the polynomial \( x^3 - 2x^2 - 5x + 6 \) decomposes into its linear factors as:
\[
(x - 1)(x - 3)(x + 2)
\]