Question : Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case :
(i) \(2x^3 + x^2 – 5x + 2\); \(1\over 2\), 1, -2
(ii) \(x^3 – 4x^2 + 5x – 2\); 2, 1, 1
Solution :
(i) Comparing the given polynomials with \(ax^3 + bx^2 + cx + d\), we get
a = 2, b = 1, c = -5 and d = 2
p(\(1\over 2\)) = 2\(({1\over 2})^3\) + \(({1\over 2})^2\) – 5(\(1\over 2\)) + 2 = \(1\over 4\) + \(1\over 4\) – \(5\over 2\) + 2 = \(0\over 4\) = 0
p(1) = 2\((1)^3\) + \((1)^2\) – 5(1) + 2 = 2 + 1 – 5 + 2 = 0
p(-2) = 2\((-2)^3\) + \((-2)^2 – 5(-2) + 2 = -16 + 16 = 0
\(\therefore\) \(1\over 2\), 1 and -2 are the zeroes of \(2x^3 + x^2 – 5x + 2\).
So, \(\alpha\) = \(1\over 2\), \(\beta\) = 1 and \(\gamma\) = -2.
Therefore, \(\alpha\) + \(\beta\) + \(\gamma\) = \(1\over 2\) + 1 + (-2) = \(-1\over 2\) = \(-b\over a\)
\(\alpha\)\(\beta\) + \(\beta\)\(\gamma\) + \(\gamma\)\(\alpha\) = (\(1\over 2\))(1) + (1)(-2) + (-2)(\(1\over 2\)) = \(-5\over 2\) = \(c\over a\)
and \(\alpha\)\(\beta\)\(\gamma\) = \(1\over 2\) \(\times\) 1 \(\times\) (-2) = \(-2\over 2\) = \(-d\over a\)
(ii) Comparing the given polynomials with \(ax^3 + bx^2 + cx + d\), we get
a = 1, b = -4, c = 5 and d = -2
p(2) = \((2)^3\) – 4\((2)^2\) + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
p(1) = \((1)^3\) – 4\((1)^2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0
\(\therefore\) 2, 1 and 1 are the zeroes of \(x^3 – 4x^2 + 5x – 2\)
So, \(\alpha\) = 2, \(\beta\) = 1 and \(\gamma\) = 1.
Therefore, \(\alpha\) + \(\beta\) + \(\gamma\) = 2 + 1 + 1 = \(-(-4)\over 1\) = \(-b\over a\)
\(\alpha\)\(\beta\) + \(\beta\)\(\gamma\) + \(\gamma\)\(\alpha\) = (2)(1) + (1)(1) + (1)(2) = \(5\over 1\) = \(c\over a\)
and \(\alpha\)\(\beta\)\(\gamma\) = 2 \(\times\) 1 \(\times\) 1 = -(-2) = \(-d\over a\)
