Finding All Zeroes of a Polynomial

Video Explanation

Question

Obtain all the zeroes of the polynomial

\[ f(x) = 2x^4 + x^3 – 14x^2 – 19x – 6, \]

if two of its zeroes are \(-2\) and \(-1\).

Solution

Step 1: Use the Given Zero \(x = -2\)

Since \(-2\) is a zero, \((x + 2)\) is a factor of the polynomial.

Dividing \(2x^4 + x^3 – 14x^2 – 19x – 6\) by \((x + 2)\), we get:

\[ 2x^4 + x^3 – 14x^2 – 19x – 6 = (x + 2)(2x^3 – 3x^2 – 8x – 3) \]

Step 2: Use the Given Zero \(x = -1\)

Since \(-1\) is a zero, \((x + 1)\) is a factor of \[ 2x^3 – 3x^2 – 8x – 3. \]

Dividing,

\[ 2x^3 – 3x^2 – 8x – 3 = (x + 1)(2x^2 – 5x – 3) \]

Step 3: Factorise the Remaining Quadratic Polynomial

\[ 2x^2 – 5x – 3 \]

\[ = (2x + 1)(x – 3) \]

Step 4: Write All the Factors

\[ f(x) = (x + 2)(x + 1)(2x + 1)(x – 3) \]

Step 5: Obtain All the Zeroes

Equating each factor to zero:

\[ x + 2 = 0 \Rightarrow x = -2 \]

\[ x + 1 = 0 \Rightarrow x = -1 \]

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

\[ x – 3 = 0 \Rightarrow x = 3 \]

Conclusion

The zeroes of the polynomial

\[ 2x^4 + x^3 – 14x^2 – 19x – 6 \]

are

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