(i) f(x) = 2x3 + x2 – 5x + 2; \(\frac{1}{2}\), 1, -2
Given,
f(x) = 2x3 + x2 – 5x + 2, where a = 2, b = 1, c = -5 and d = 2
For x = \(\frac{1}{2}\)
f(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))3 + (\(\frac{1}{2}\))2 – 5(\(\frac{1}{2}\)) + 2
= \(\frac{1}{4}\) + \(\frac{1}{4}\) – \(\frac{5}{2}\) + 2
= 0
⇒ f(\(\frac{1}{2}\)) = 0, hence x = \(\frac{1}{2}\) is a root of the given polynomial.
For x = 1
f(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2
= 0
⇒ f(1) = 0, hence x = 1 is also a root of the given polynomial.
For x = -2
f(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
= -16 + 4 + 10 + 2
= 0
⇒ f(-2) = 0, hence x = -2 is also a root of the given polynomial.
Now,
Sum of zeros = -b/a
\(\frac{1}{2}\) + 1 – 2 = – (1)/2
-\(\frac{1}{2}\) = -\(\frac{1}{2}\)
Sum of the products of the zeros taken two at a time = c/a
(\(\frac{1}{2}\) x 1) + (1 x -2) + (\(\frac{1}{2}\) x -2) = -\(\frac{5}{2}\)
\(\frac{1}{2}\) – 2 + (-1) = -\(\frac{5}{2}\)
-\(\frac{5}{2}\) = -\(\frac{5}{2}\)
Product of zeros = – \(\frac{d}{a}\)
\(\frac{1}{2}\) x 1 x (– 2) = -(2)/2
-1 = -1
Hence, the relationship between the zeros and coefficients is verified.
(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1
Given,
g(x) = x3 – 4x2 + 5x – 2, where a = 1, b = -4, c = 5 and d = -2
For x = 2
g(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 0
⇒ f(2) = 0, hence x = 2 is a root of the given polynomial.
For x = 1
g(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
⇒ g(1) = 0, hence x = 1 is also a root of the given polynomial.
Now, Sum of zeros = -\(\frac{b}{a}\)
1 + 1 + 2 = – (-4)/1
4 = 4
Sum of the products of the zeros taken two at a time = \(\frac{c}{a}\)
(1 x 1) + (1 x 2) + (2 x 1) = \(\frac{5}{1}\)
1 + 2 + 2 = 5
5 = 5
Product of zeros = – \(\frac{d}{a}\)
1 x 1 x 2 = -(-2)/1
2 = 2
Hence, the relationship between the zeros and coefficients is verified.
