For which values of \(a\) and \(b\), are the zeroes of \(q(x)=x^3+2x^2+a\) also the zeroes of \(p(x)=x^5-x^4-4x^3+3x^2+3x+b\)? Which zeroes of \(p(x)\) are not zeroes of \(q(x)\)?
\(a = -1\), \(b = -2\).
Then \(q(x) = x^3 + 2x^2 - 1 = (x + 1)(x^2 + x - 1)\).
And \(p(x) = q(x)\,(x^2 - 3x + 2) = q(x)(x - 1)(x - 2)\).
So, the zeroes common to both are the three zeroes of \(q\): \(x = -1\), \(x = \dfrac{-1 \pm \sqrt{5}}{2}\).
The zeroes of \(p\) that are not zeroes of \(q\) are \(x = 1\) and \(x = 2\).
Step 1: Require exact divisibility.
For the zeroes of \(q\) to be zeroes of \(p\), the remainder on dividing \(p\) by \(q\) must be \(0\).
Step 2: Compute the remainder symbolically.
Divide \(p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b\) by \(q(x) = x^3 + 2x^2 + a\).
The remainder has the form:
\(-(a+1)x^2 + (3a+3)x + (-2a + b)\).
Step 3: Set each coefficient to zero.
From \(-(a+1) = 0\Rightarrow a = -1\).
From \(3a + 3 = 0\Rightarrow a = -1\) (consistent).
From \(-2a + b = 0\Rightarrow b = 2a = -2\).
Step 4: Factor with these values.
\(q(x) = x^3 + 2x^2 - 1 = (x + 1)(x^2 + x - 1)\).
\(p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x - 2\)
\(\quad\;= q(x)\,(x^2 - 3x + 2) = q(x)(x - 1)(x - 2)\).
Step 5: Read off the zeroes.
Common zeroes (from \(q\)): \(x = -1\), \(x = \dfrac{-1 \pm \sqrt{5}}{2}\).
Additional zeroes of \(p\): \(x = 1\) and \(x = 2\).