Question

Find the cubic polynomial whose zeroes are 3, 5 and -2.

Answer 1

Hint: The question is related to polynomial. We have to make the cubic polynomial using the zeros given in the question. Use the sum of zeroes, product of the zeroes and sum of the product of the zero’s formula. Zeroes of the cubic polynomials are $\alpha ,\beta ,\gamma $. Here $\alpha $ is equal to 3 ,$\beta $ is equal to 5 and $\gamma $ is equal to -2. In the cubic polynomial the coefficient of ${x^3}$ is a, coefficient of ${x^2}$ is b, coefficient of $x$ is c and the consent term is d. use the formula to get your cubic polynomial equation.

Complete step by step solution:
Given that the zeroes of the cubic polynomial are 3, 5 and-2 that means $(x + 3),(x + 5),(x - 2)$
We know that the zeroes of cubic polynomial is denoted by $\alpha ,\beta ,\gamma $
Here we know that the sum of the zeroes is
${\text{sum of zeroes = }}\dfrac{{{\text{coefficient of }}{{\text{x}}^2}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
$\alpha + \beta + \gamma = \dfrac{{ - b}}{a}$
Putting the vales of the zero in above formula
$3 + 5 - 2 = \dfrac{{ - b}}{c}$
Solve the left-hand side
$ \Rightarrow 6 = \dfrac{{ - b}}{c}$
Product of the zeroes is
${\text{product of the zeroes = }}\dfrac{{{\text{consent term}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
$\alpha \times \beta \times \gamma = \dfrac{{ - d}}{a}$
Putting the values of the zeroes
$3 \times 5 \times ( - 2) = \dfrac{{ - d}}{a}$
Solve the left-hand side
$ \Rightarrow - 30 = \dfrac{{ - d}}{a}$
Sum of the product of zeroes is
${\text{sum of product of zeroes = }}\dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
$\alpha \times \beta + \beta \times \gamma + \alpha \times \gamma = \dfrac{c}{a}$
Putting the values of zeroes
$3 \times 5 + 5 \times ( - 2) + 3 \times ( - 2) = \dfrac{c}{a}$
Solve the left-hand side
$ \Rightarrow 15 - 10 - 6 = \dfrac{c}{a}$
$ \Rightarrow - 1 = \dfrac{c}{a}$
On comparing the above solutions, we get the cubic polynomial
$a = 1,b = - 6,c - 1,d = 30$
Equation
$\therefore {x^3} - 6{x^2} - x + 30 = 0$

Hence the cubic polynomial equation is ${x^3} - 6{x^2} - x + 30 = 0$.

Note: Here in this question students mostly make the mistake in the sigh of plus and minus as -b means you have to write the value of b with the sigh of – in the equation. Solve the solution step by step. You must know the formula of solving zeroes.
${\text{sum of zeroes = }}\dfrac{{{\text{coefficient of }}{{\text{x}}^2}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{product of the zeroes = }}\dfrac{{{\text{consent term}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{sum of product of zeroes = }}\dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$