Finding All Zeroes of a Polynomial

Video Explanation

Question

Find all the zeroes of the polynomial

\[ f(x) = 2x^4 – 9x^3 + 5x^2 + 3x – 1, \]

if two of its zeroes are \(2+\sqrt{3}\) and \(2-\sqrt{3}\).

Solution

Step 1: Form the Quadratic Factor from the Given Zeroes

Since the given zeroes are conjugates,

\[ (x-(2+\sqrt{3}))(x-(2-\sqrt{3})) \]

\[ = (x-2)^2 – 3 = x^2 – 4x + 1 \]

Hence, \(x^2 – 4x + 1\) is a factor of the given polynomial.

Step 2: Divide the Polynomial by \(x^2 – 4x + 1\)

Dividing

\[ 2x^4 – 9x^3 + 5x^2 + 3x – 1 \]

by

\[ x^2 – 4x + 1, \]

we get:

\[ 2x^4 – 9x^3 + 5x^2 + 3x – 1 = (x^2 – 4x + 1)(2x^2 – x – 1) \]

Step 3: Factorise the Remaining Quadratic Polynomial

\[ 2x^2 – x – 1 \]

\[ = (2x + 1)(x – 1) \]

Step 4: Write the Complete Factorisation

\[ f(x) = (x^2 – 4x + 1)(2x + 1)(x – 1) \]

Step 5: Obtain All the Zeroes

Equating each factor to zero:

\[ x^2 – 4x + 1 = 0 \Rightarrow x = 2 \pm \sqrt{3} \]

\[ 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \]

\[ x – 1 = 0 \Rightarrow x = 1 \]

Conclusion

The zeroes of the polynomial

\[ 2x^4 – 9x^3 + 5x^2 + 3x – 1 \]

are

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