Question

If f, of, x, equals, x, cubed, minus, 11, x, squared, plus, 36, x, minus, 36f(x)=x
3
−11x
2
+36x−36 and x, minus, 2x−2 is a factor of f, of, xf(x), then find all of the zeros of f, of, xf(x) algebraically.

Answer 1

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 \( x - 2 \):

1. Divide the leading term: \( \frac{x^3}{x} = x^2 \).
2. Multiply \( x^2 \) by \( x - 2 \): \( x^2(x - 2) = x^3 - 2x^2 \).
3. Subtract: \((x^3 - 11x^2 + 36x - 36) - (x^3 - 2x^2) = -11x^2 + 2x^2 + 36x - 36 = -9x^2 + 36x - 36\).
4. Repeat: Divide the leading term: \( \frac{-9x^2}{x} = -9x \).
5. Multiply: \( -9x(x - 2) = -9x^2 + 18x \).
6. Subtract: \((-9x^2 + 36x - 36) - (-9x^2 + 18x) = 36x - 18x - 36 = 18x - 36\).
7. Repeat: Divide leading term: \( \frac{18x}{x} = 18 \).
8. Multiply: \( 18(x - 2) = 18x - 36 \).
9. 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:

1. \( x = \frac{9 + 3}{2} = \frac{12}{2} = 6 \)
2. \( x = \frac{9 - 3}{2} = \frac{6}{2} = 3 \)

Thus, the zeros of \( f(x) = 0 \) are:

\[
x = 2, \quad x = 3, \quad x = 6
\]

In conclusion, the zeros of \( f(x) \) are \( \boxed{2, 3, 6} \).