Question

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a
time, and the product of its zeroes as 2, -7, -14 respectively.​

Answer 1

☆ ANSWER:-

Your Answer is $\tt x^3-2x^2-7x+14.$

Given:-

• Sum of zeroes = 2.

• Sum of the zeroes taken two at a time = -7.

• Product of zeroes = -14.

To Find:-

• The Cubic Polynomial.

Concepts Used:-

1. Standard form of of a cubic polynomial,

$\tt\leadsto a {x}^{3} + b {x}^{2} + cx + d.$

2. Sum of zeroes in cubic polynomial = $\frac{-b}{a}.$

3. Product of zeroes in a cubic polynomial = $\frac{-d}{a}.$

4. Sum of zeroes taken two at a time in a cubic polynomial = $\frac{c}{a}.$

So, Here:-

$\\ \tt\bullet \:\: Sum\:\:Of\:\:Zeroes = 2 = \frac{-b}{a}. \\ \\ \tt\leadsto \frac{-b}{a} = 2. \\ \\ \\ \tt\bullet \:\: Product\:\:Of\:\:Zeroes = -14 = \frac{-d}{a}. \\ \\ \tt\leadsto \frac{-d}{a} = -14. \\ \\ \\ \tt\bullet \:\: Sum\:\:Of\:\:Zeroes\:\:taken\:\:two\:\:at\:\:a\:\:time = -7 = \frac{c}{a}. \\ \\ \tt\leadsto \frac{c}{a} = -7.$

Therefore By Comparing we get:-

$\sf \bullet \:a = 1. \\ \\ \sf\bullet \: -b = 2 \implies b = -2. \\ \\ \sf\bullet\: c = -7. \\ \\ \sf\bullet\:\cancel{-}d = \cancel{-}14. \implies d = 14.$

So The required cubic polynomial is,

$\tt \mapsto \: a {x}^{3} + bx {}^{2} + cx + d. \\ \\ \tt = 1(x) {}^{3} + ( - 2) {x}^{2} + ( - 7)x + (14) .\\ \\ \tt = {x}^{3} - 2 {x}^{2} - 7x + 14.$