Question

# We are given that $\alpha ,\beta \,\& \gamma$ are the zeroes of cubic polynomial $P(x) = a{x^3} + b{x^3} + cx + d,(a \ne 0)$ then product of their zeroes $\left[ {\alpha .\beta .\gamma } \right] = .....$

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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:

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 $\Rightarrow \alpha \beta = \dfrac{c}{a}$