Question

15. In a cubic polynomial, sum of the zeroes is 9 and product of zeroes
is 15. If zeroes are in A.P, then find that cubic polynomial.​

Answer 1

This can be Formed cubic polynomial by cubic polynomial formula in the the given image

Answer 2

$x^3-9x^2+23x-15$

Step-by-step explanation:

Let $\alpha,\beta$ and $\gamma$ are zeroes of cubic polynomial.

According to question

$\alpha+\beta+\gamma=9$

$\alpha+\gamma=9-\beta$

$\alpha \beta \gamma=15$

$\alpha,\beta$ and $\gamma$ are in A.P

Therefore,

$\beta-\alpha=\gamma-\beta$

$\beta+\beta=\gamma+\alpha$

$2\beta=\gamma+\alpha$

Substitute the values

$2\beta=9-\beta$

$2\beta+\beta=9$

$3\beta=9$

$\beta=\frac{9}{3}=3$

Substitute the values

$\alpha+\gamma=9-3=6$

$3\alpha\gamma=15$

$\alpha\gamma=\frac{15}{3}=5$

$\gamma=\frac{5}{\alpha}$

Substitute the values

$\alpha+\frac{5}{\alpha}=6$

$\frac{\alpha^2+5}{\alpha}=6$

$\alpha^2+5=6\alpha$

$\alpha^2-6\alpha+5=0$

$\alpha^2-5\alpha-\alpha+5=0$

$\alpha(\alpha-5)-1(\alpha-5)=0$

$(\alpha-5)(\alpha-1)=0$

$\alpha-5=0\implies \alpha=5$

$\alpha-1=0\implies \alpha=1$

Substitute the value of $\alpha=5$

$\gamma=\frac{5}{5}=1$

When $\alpha=1$

$\gamma=\frac{5}{1}=5$

General  equation of cubic polynomial

$x^3-(\alpha+\beta+\gamma)x^2+(\alpha \beta+\beta\gamma+\gamma\alpha)x-\alpha\beta \gamma$

Substitute the values$\alpha=5,\beta=3,\gamma=1$

$x^3-9x^2+(5(3)+3(1)+1(5))-15$

$x^3-9x^2+23x-15=0$

Substitute the values$\alpha=1,\beta=3,\gamma=5$

$x^3-9x^2+(1(3)+3(5)+5(1))-15$

$x^3-9x^2+23x-15$

#Learn more:

https://brainly.in/question/15858623:Answered by JHP