Answer
Verified
105.3k+ views
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.
${\text{sum of zeroes = }}\dfrac{{{\text{coefficient of }}{{\text{x}}^2}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{product of the zeroes = }}\dfrac{{{\text{consent term}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{sum of product of zeroes = }}\dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
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.
${\text{sum of zeroes = }}\dfrac{{{\text{coefficient of }}{{\text{x}}^2}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{product of the zeroes = }}\dfrac{{{\text{consent term}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
${\text{sum of product of zeroes = }}\dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main