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Question: If we are given that \(\alpha ,\beta ,\gamma \) are the zeroes of cubic polynomial \(3{{x}^{3}}-2{{x...

If we are given that α,β,γ\alpha ,\beta ,\gamma are the zeroes of cubic polynomial 3x3−2x2+5x−6=03{{x}^{3}}-2{{x}^{2}}+5x-6=0 then find
( a ) α+β+γ\alpha +\beta +\gamma
( b ) α⋅β+β⋅γ+γ⋅α\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha

Explanation

Solution

We will use the results based on of relationship between roots of cubic equation to evaluate the values of α+β+γ\alpha +\beta +\gamma and α⋅β+β⋅γ+γ⋅α\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha which are given by the relation of coefficients of cubic equation such as α+β+γ=−ba\alpha +\beta +\gamma =-\dfrac{b}{a} and α⋅β+β⋅γ+γ⋅α=ca\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha =\dfrac{c}{a} for cubic equation ax3+bx2+cx+d=0a{{x}^{3}}+b{{x}^{2}}+cx+d=0 where a≠0.a\ne 0.

Complete step-by-step solution:
Now, firstly we will find the coefficients of x3{{x}^{3}} ,x2{{x}^{2}} ,xx and constant d from the given polynomial 3x3−2x2+5x−6=03{{x}^{3}}-2{{x}^{2}}+5x-6=0 by comparing it with the general form of cubic polynomial which is expressed as ax3+bx2+cx+d=0a{{x}^{3}}+b{{x}^{2}}+cx+d=0.
On comparing given polynomial with general form of cubic polynomial, we get coefficients of cubic polynomial 3x3−2x2+5x−6=03{{x}^{3}}-2{{x}^{2}}+5x-6=0equals to,
a = 3, b = -2, c = 5, d = -6
Now,
( a ) Here, we know that α+β+γ=−ba\alpha +\beta +\gamma =-\dfrac{b}{a}……( i ),
So, we can obtain the value of α+β+γ\alpha +\beta +\gamma easily by substituting the values of b and a in an equation ( i )
Substituting values of a = 3 and b = -2 inα+β+γ=−ba\alpha +\beta +\gamma =-\dfrac{b}{a}, we get
α+β+γ=−(−2)3\alpha +\beta +\gamma =-\dfrac{(-2)}{3}
On simplifying signs, we get
α+β+γ=23\alpha +\beta +\gamma =\dfrac{2}{3}
( b ) Here, we know that α⋅β+β⋅γ+γ⋅α=ca\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha =\dfrac{c}{a}…..( ii ),
So, we can obtain the value of α⋅β+β⋅γ+γ⋅α\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha easily by substituting the values of c and a in an equation ( ii )
Substituting values of a = 3 and c= 5 in α⋅β+β⋅γ+γ⋅α=ca\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha =\dfrac{c}{a}, we get
α⋅β+β⋅γ+γ⋅α=53\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha =\dfrac{5}{3}.
Hence, the values of α+β+γ\alpha +\beta +\gamma and α⋅β+β⋅γ+γ⋅α\alpha \cdot \beta +\beta \cdot \gamma +\gamma \cdot \alpha are equals to 23\dfrac{2}{3} and 53\dfrac{5}{3} respectively .

Note: Remember these formulae as they are very helpful in solving questions. While calculating the coefficients of a cubic equation, try to avoid signs error as this makes the answer incorrect. Simplification of signs should be done carefully.