Find \(k\) so that \(x^2 + 2x + k\) is a factor of \(2x^4 + x^3 - 14x^2 + 5x + 6\). Also find all zeroes of the two polynomials.
\(k = -3\).
Zeroes of \(x^2 + 2x - 3\): \(x = 1\), \(x = -3\).
Zeroes of \(2x^4 + x^3 - 14x^2 + 5x + 6\): \(x = -\dfrac{1}{2}\), \(x = 2\), \(x = 1\), \(x = -3\).
Step 1: Assume divisibility and compare coefficients.
Let \(2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k)(2x^2 + ax + b)\).
Step 2: Expand the right-hand side.
After expanding and collecting like terms, equate coefficients of \(x^3, x^2, x\) and constant.
This gives the system:
\(a + 3 = 0\),
\(2a + b + 2k + 14 = 0\),
\(ak + 2b - 5 = 0\),
\(bk - 6 = 0\).
Step 3: Solve step by step.
From \(a + 3 = 0\Rightarrow a = -3\).
Then \(2(-3) + b + 2k + 14 = 0 \Rightarrow b + 2k + 8 = 0\Rightarrow b = -2k - 8\).
Next \((-3)k + 2(-2k - 8) - 5 = 0 \Rightarrow -7k - 21 = 0\Rightarrow k = -3\).
Hence \(b = -2(-3) - 8 = -2\).
The factorisation is:
\((x^2 + 2x - 3)(2x^2 - 3x - 2)\).
Step 4: Factor completely and read the roots.
\(x^2 + 2x - 3 = (x - 1)(x + 3)\Rightarrow x = 1,\, -3\).
\(2x^2 - 3x - 2 = (2x + 1)(x - 2)\Rightarrow x = -\dfrac{1}{2},\, 2\).