To find the zeros of the polynomial \( f(x) = x^3 - 11x^2 + 36x - 36 \) given that \( x - 2 \) is a factor, we can perform polynomial division or use the Factor Theorem.
Since \( x - 2 \) is a factor, it means that \( f(2) = 0 \). Let's verify:
\[ f(2) = 2^3 - 11(2^2) + 36(2) - 36 \] \[ = 8 - 44 + 72 - 36 \] \[ = 8 - 44 + 72 - 36 = 0 \]
Now let's perform polynomial long division of \( f(x) \) by \( x - 2 \):
- Divide the leading term: \( \frac{x^3}{x} = x^2 \).
- Multiply \( x^2 \) by \( x - 2 \): \( x^2(x - 2) = x^3 - 2x^2 \).
- Subtract: \((x^3 - 11x^2 + 36x - 36) - (x^3 - 2x^2) = -11x^2 + 2x^2 + 36x - 36 = -9x^2 + 36x - 36\).
- Repeat: Divide the leading term: \( \frac{-9x^2}{x} = -9x \).
- Multiply: \( -9x(x - 2) = -9x^2 + 18x \).
- Subtract: \((-9x^2 + 36x - 36) - (-9x^2 + 18x) = 36x - 18x - 36 = 18x - 36\).
- Repeat: Divide leading term: \( \frac{18x}{x} = 18 \).
- Multiply: \( 18(x - 2) = 18x - 36 \).
- Subtract: \((18x - 36) - (18x - 36) = 0\).
The result of the division is \( x^2 - 9x + 18 \). So, we can express \( f(x) \) as:
\[ f(x) = (x - 2)(x^2 - 9x + 18) \]
Next, we need to find the zeros of \( x^2 - 9x + 18 \) by using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -9 \), and \( c = 18 \).
Calculate the discriminant:
\[ b^2 - 4ac = (-9)^2 - 4(1)(18) = 81 - 72 = 9 \]
Now substitute into the quadratic formula:
\[ x = \frac{9 \pm \sqrt{9}}{2(1)} = \frac{9 \pm 3}{2} \]
This gives us two solutions:
- \( x = \frac{9 + 3}{2} = \frac{12}{2} = 6 \)
- \( x = \frac{9 - 3}{2} = \frac{6}{2} = 3 \)
Thus, the zeros of \( f(x) = 0 \) are:
\[ x = 2, \quad x = 3, \quad x = 6 \]
In conclusion, the zeros of \( f(x) \) are \( \boxed{2, 3, 6} \).
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