Find a cubic polynomial whose zeroes are \(\frac{1}2\) , 1 and –3.

1.	Find a cubic polynomial whose zeroes are \(\frac{1}2\) , 1 and –3.
Answer» If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1) Let a = \(\frac{1}{2}\), b = 1 and c = –3 Substituting the values in (1), we get x³ – \((\frac{1}3+1-3)\) x² + \((\frac{1}3-3-\frac{3}2)\)x - \((\frac{-3}2)\) ⇒ x³ – \((\frac{-3}2)\)x² - 4x = \(\frac{3}2\) ⇒ 2x³ + 3x² – 8x + 3

Answer»

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as

x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1)

Let a = \(\frac{1}{2}\), b = 1 and c = –3

Substituting the values in (1), we get

x³ – \((\frac{1}3+1-3)\) x² + \((\frac{1}3-3-\frac{3}2)\)x - \((\frac{-3}2)\)

⇒ x³ – \((\frac{-3}2)\)x² - 4x = \(\frac{3}2\)

⇒ 2x³ + 3x² – 8x + 3

Discussion