Finding All Zeroes of a Cubic Polynomial
Video Explanation
Question
Given that \(x-\sqrt{5}\) is a factor of the cubic polynomial
\[ f(x) = x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5}, \]
find all the zeroes of the polynomial.
Solution
Step 1: Use the Conjugate Factor Theorem
Since the coefficients of the polynomial are real and \(x-\sqrt{5}\) is a factor, its conjugate \(x+\sqrt{5}\) is also a factor.
Hence,
\[ (x-\sqrt{5})(x+\sqrt{5}) = x^2 – 5 \]
is a factor of the given polynomial.
Step 2: Divide the Polynomial by \(x^2 – 5\)
Dividing
\[ x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5} \]
by
\[ x^2 – 5, \]
we get:
\[ x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5} = (x^2 – 5)(x – 3\sqrt{5}) \]
Step 3: Obtain All the Zeroes
Equating each factor to zero:
\[ x^2 – 5 = 0 \Rightarrow x = \pm \sqrt{5} \]
\[ x – 3\sqrt{5} = 0 \Rightarrow x = 3\sqrt{5} \]
Conclusion
The zeroes of the given polynomial
\[ x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5} \]
are
\[ \boxed{-\sqrt{5},\; \sqrt{5},\; 3\sqrt{5}} \]