Question

Decompose the polynomial x3−2x2−5x+6
into its linear factors, given that x−1
is a factor. (1 point)

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 to divide the polynomial by \( x - 1 \).

1. Perform the polynomial long division of \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \).

\[
\begin{array}{r|rrr}
& 1 & -1 & -5 & +6 \\
x-1 & 1 & -2 & -5 & +6 \\
& & 1 & -1 & -6 \\
\hline
& 1 & -1 & -6& 0 \\
\end{array}
\]

The result of this division is \( x^2 - x - 6 \), and the remainder is 0, confirming that \( x - 1 \) is indeed a factor.

2. Next, we need to factor \( x^2 - x - 6 \). We can do this by finding two numbers that multiply to \(-6\) (the constant term) and add up to \(-1\) (the coefficient of \( x \)). The numbers \(-3\) and \(2\) satisfy this.

Thus, we can factor \( x^2 - x - 6 \) as:

\[
(x - 3)(x + 2)
\]

3. Now we can express the original polynomial \( x^3 - 2x^2 - 5x + 6 \) in factored form:

\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)
\]

Therefore, the complete factorization of the polynomial \( x^3 - 2x^2 - 5x + 6 \) into its linear factors is:

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