Sure, let’s find the zeros of the given polynomial [tex]\( P(x) = x^3 - 6x^2 + 11x - 6 \)[/tex]. Here's how we can approach this step-by-step:
### Step 1: Understand the Polynomial
We are given the polynomial [tex]\( P(x) = x^3 - 6x^2 + 11x - 6 \)[/tex]. Our goal is to find the values of [tex]\( x \)[/tex] for which [tex]\( P(x) = 0 \)[/tex].
### Step 2: Set the Polynomial Equal to Zero
To find the zeros of [tex]\( P(x) \)[/tex], we set the polynomial equal to zero:
[tex]\[ x^3 - 6x^2 + 11x - 6 = 0 \][/tex]
### Step 3: Factor the Polynomial
One way to solve a cubic equation is by factoring. We can first try to guess a rational root using the Rational Root Theorem, which suggests that any rational root is a factor of the constant term divided by the leading coefficient. In this polynomial, the constant term is [tex]\(-6\)[/tex] and the leading coefficient is [tex]\(1\)[/tex], so the possible rational roots are [tex]\( \pm 1, \pm 2, \pm 3, \pm 6 \)[/tex].
Let's test these potential roots to see if they satisfy the equation:
1. Test [tex]\( x = 1 \)[/tex]:
[tex]\[ 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \][/tex]
Since [tex]\( P(1) = 0 \)[/tex], [tex]\( x = 1 \)[/tex] is a root.
### Step 4: Use Synthetic Division
Having found [tex]\( x = 1 \)[/tex] as a root, we can now factor [tex]\( x - 1 \)[/tex] out of the polynomial. We perform synthetic division to divide [tex]\( x^3 - 6x^2 + 11x - 6 \)[/tex] by [tex]\( x - 1 \)[/tex].
Performing the synthetic division:
[tex]\[
\begin{array}{r|rrrr}
1 & 1 & -6 & 11 & -6 \\
& & 1 & -5 & 6 \\
\hline
& 1 & -5 & 6 & 0 \\
\end{array}
\][/tex]
Resulting polynomial: [tex]\( x^2 - 5x + 6 \)[/tex]
### Step 5: Factor the Quadratic Polynomial
Next, we factor the quadratic polynomial [tex]\( x^2 - 5x + 6 \)[/tex]:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
### Step 6: Combine All Factors
Now we have factored the original polynomial completely:
[tex]\[ P(x) = (x - 1)(x - 2)(x - 3) \][/tex]
### Step 7: Identify All Zeros
Setting each factor equal to zero gives us the zeros:
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
Thus, the zeros of the polynomial [tex]\( P(x) = x^3 - 6x^2 + 11x - 6 \)[/tex] are:
[tex]\[ x = 1, x = 2, \text{ and } x = 3 \][/tex]
### Conclusion
The polynomial [tex]\( P(x) = x^3 - 6x^2 + 11x - 6 \)[/tex] has zeros at [tex]\( x = 1 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 3 \)[/tex].
