For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i). \(-\dfrac{8}{3}\), \(\dfrac{4}{3}\)
(ii). \(\dfrac{21}{8}\), \(\dfrac{5}{16}\)
(iii). \(-2\sqrt{3}\), \(-9\)
(iv). \(-\dfrac{3}{2\sqrt{5}}\), \(-\dfrac{1}{2}\)
(i) Polynomial: \(3x^2 + 8x + 4\); Zeroes: \(x=-2\), \(x=-\dfrac{2}{3}\).
(ii) Polynomial: \(16x^2 - 42x + 5\); Zeroes: \(x=\dfrac{5}{2}\), \(x=\dfrac{1}{8}\).
(iii) Polynomial: \(x^2 + 2\sqrt{3}x - 9\); Zeroes: \(x=\sqrt{3}\), \(x=-3\sqrt{3}\).
(iv) Polynomial: \(2\sqrt{5}x^2 + 3x - \sqrt{5}\); Zeroes: \(x=\dfrac{1}{\sqrt{5}}\), \(x=-\dfrac{\sqrt{5}}{2}\).
Given that the zeroes of \(x^3-6x^2+3x+10\) are of the form \(a,\, a+b,\, a+2b\), find \(a\), \(b\) and the zeroes.
\(a=5,\; b=-3\) (equivalently \(a=-1,\; b=3\)); Zeroes: \(\{5,\,2,\,-1\}\).
Given that \(\sqrt{2}\) is a zero of \(6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}\), find the other two zeroes.
Other zeroes: \(x = -\dfrac{\sqrt{2}}{2}\), \(x = -\dfrac{2\sqrt{2}}{3}\).
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 \(2x^4 + x^3 - 14x^2 + 5x + 6\): \(x = -\dfrac{1}{2}\), \(x = 2\), \(x = 1\), \(x = -3\).
Given that \(x - 5\) is a factor of \(x^3 - 3\sqrt{2}\,x^2 + 13x - 3\sqrt{5}\), find all zeroes.
\(x = 5\), \(x = \sqrt{2} + \sqrt{5}\), \(x = \sqrt{2} - \sqrt{5}\).
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 \(x = 2\).