Finding All Zeroes of a Polynomial

Video Explanation

Question

Obtain all the zeroes of the polynomial

\[ f(x) = x^4 – 3x^3 – x^2 + 9x – 6, \]

if two of its zeroes are \( -\sqrt{3} \) and \( \sqrt{3} \).

Solution

Step 1: Form the Quadratic Factor from Given Zeroes

Since the zeroes are \( -\sqrt{3} \) and \( \sqrt{3} \),

\[ (x – \sqrt{3})(x + \sqrt{3}) = x^2 – 3 \]

Hence, \(x^2 – 3\) is a factor of the given polynomial.

Step 2: Divide the Polynomial by \(x^2 – 3\)

Dividing

\[ x^4 – 3x^3 – x^2 + 9x – 6 \]

by

\[ x^2 – 3, \]

we get:

\[ x^4 – 3x^3 – x^2 + 9x – 6 = (x^2 – 3)(x^2 – 3x + 2) \]

Step 3: Factorise the Remaining Quadratic Polynomial

\[ x^2 – 3x + 2 \]

\[ = (x – 1)(x – 2) \]

Step 4: Write the Complete Factorisation

\[ f(x) = (x^2 – 3)(x – 1)(x – 2) \]

Step 5: Obtain All the Zeroes

Equating each factor to zero:

\[ x^2 – 3 = 0 \Rightarrow x = \pm \sqrt{3} \]

\[ x – 1 = 0 \Rightarrow x = 1 \]

\[ x – 2 = 0 \Rightarrow x = 2 \]

Conclusion

The zeroes of the polynomial

\[ x^4 – 3x^3 – x^2 + 9x – 6 \]

are

\[ \boxed{-\sqrt{3},\; \sqrt{3},\; 1,\; 2} \]