Question

If , are the zeroes of the polynomial x^3-6x^2+11x-6, then find the values of sum of the product of zeroes taken two at a time , product of zeroes

Answer 1

Answer:

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Sum = 6

product = 6

Step-by-step explanation:

Let the $\sf{\alpha\:\beta\:\gamma}$ be the zeroes of the polynomial

Here p(x) = $\sf{x^3\:-\:6x^2\:+\:11x\:-\:6}$ is of the form $\sf{ax^3\:+\:bx^2\:+\:cx\:+\:d}$

Therefore,

➝a = 1

➝b = -6

➝c = 11

➝d = -6

Here, it is.given that product of two zeroes is 3.

➝$\sf{\alpha\:\beta\:=\:3}$.,.,.,.,.,.,.,eq^n (i)

Now,

product of three zeroes.

➝$\sf{\alpha\beta\gamma\:=\:\frac{-d}{a}}$

➝$\sf{\alpha\beta\gamma\:=\:\frac{-(-6)}{1}}$

➝$\sf{\alpha\beta\gamma\:=\:6}$

➝$\sf{\gamma\:=\:\frac{6}{\alpha\beta}}$

➝$\sf{\gamma\:=\:\frac{6}{3}\:(from\:eq^n\:(i))}$

➝$\sf{\gamma\:=\:\cancel\frac{6}{3}\:2}$

➝$\sf{\gamma\:=\:2}$

so, now we have all three zeroes.

Now find sum,

➝$\sf{Sum\:=\:\frac{-b}{a}}$

➝$\sf{\alpha\:+\:\beta\:+\:\gamma\:=\:\frac{-(-6)}{1}}$

Hence,

➝Sum($\sf{\alpha\:+\:\beta\:+\:\gamma}$) = 6

➝product($\sf{\alpha\beta\gamma}$) = 6

$\Large\mathcal\pink{Hope \: it \: helps \: you\:friend}$