Question

find a cubic polynomial with the sum of zero sum of the product of its zeros taken two at a time and the product of the zeros are 8,9 and - 18 respectively​

Answer 1

Step-by-step explanation:

if

$\alpha \beta \: and \: \gamma$

are the zeros of a cubic polynomial f (x)

f(x)=k (

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

where k is any non zero real number

$\alpha + \beta + \gamma = 2$

$\alpha \beta + \beta \gamma + \beta \gamma \alpha - 7 \: and \: \alpha \beta \gamma = - 14$

f (x)=k (x^3 -2x^2 - 7x +14)

where k is any non zero real number

Answer 2

The required polynomial is $P(x)=a(x^3-8x^2+9x+18)$ where, a is constant.

Step-by-step explanation:

If α,β,γ are three zeroes of the cubic polynomial

$P(x)=ax^3+bx^2+cx+d$         .... (1)

then,

$\alpha+\beta+\gamma=-\dfrac{b}{a}$

$\alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}$

$\alpha\beta\gamma=-\dfrac{d}{a}$

It is given that the sum of zero sum of the product of its zeros taken two at a time and the product of the zeros are 8,9 and - 18 respectively​.

$8=-\dfrac{b}{a}\Rightarrow b=-8a$

$9=\dfrac{c}{a}\Rightarrow c=9a$

$-18=-\dfrac{d}{a}\Rightarrow d=18a$

Substitute the value of b,c, and d inn equation (1).

$P(x)=ax^3-8ax^2+9ax+18a$

$P(x)=a(x^3-8x^2+9x+18)$

where, a is constant.

#Learn more

If p(x)=1/3x^2-5x+3/2 then find the sum and product for all its zeroes .

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