Answer
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Hint: To solve this type of problem first we will write general equation of cubic polynomial which is \[a{x^3} + b{x^2} + cx + d = 0\] and then we will make coefficient of \[x^3\] as 1 using division operation. Then we will write a cubic polynomial equation using zeros in the form of factors and then simplify and compare both equations to get the desired result.
Complete step-by- step solution:
Complete step-by- step solution:
General cubic polynomial can be written as -
\[p(x) = a{x^3} + b{x^2} + cx + d(a \ne 0)\]
Now cubic equation can be written as-
\[ \Rightarrow a{x^3} + b{x^2} + cx + d = 0\]
\[ \Rightarrow {x^3} + \dfrac{b}{a}{x^2} + \dfrac{c}{a}x + \dfrac{d}{a} = 0....(3)\]
Now,
If \[\alpha ,\beta ,\gamma \] are zeros then
\[(x - \alpha )(x - \beta )(x - \gamma ) = 0\]
On multiplying first two, we get:
\[({x^2} - (\alpha + \beta )x + \alpha \beta )(x - \gamma ) = 0\]
On multiplying the result with \[(x - \gamma )\], we get:
\[{x^3} - (\alpha + \beta ){x^2} + \alpha \beta x - \gamma {x^2} + \gamma (\alpha + \beta )x - \alpha \beta \gamma = 0\]
On simplifying, we get:
\[{x^3} - (\alpha + \beta + \gamma ){x^2} + (\alpha \beta + \beta \gamma + \gamma \alpha ) x - \alpha \beta \gamma = 0..........(4)\]
Now compare of equation (3) and (4), product of roots \[\alpha \beta \gamma = - \dfrac{d}{a}\]
Note: In quadratic equation like this,
\[a{x^2} + bx + c = 0(a \ne 0)\]
On dividing the equation throughout by ‘a’, we have:
\[ \Rightarrow {x^2} + \dfrac{b}{a} \times x + \dfrac{c}{a} = 0.......(1)\]
Now if \[\alpha \] and \[\beta \] zeroes of the quadratic equation, then:
\[(x - \alpha )(x - \beta ) = 0\]
\[{x^2} - \alpha x - \beta x + \alpha b = 0\]
\[{x^2} - (\alpha + \beta )x + \alpha b = 0....(2)\]
From equation 1 & 2, we have:
Sum of zeroes \[ \Rightarrow \alpha + \beta = - \dfrac{b}{a}\]
Product of zeroes \[ \Rightarrow \alpha \beta = \dfrac{c}{a}\]
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