The zeros of p(x) = 3x^2 - x - 4 are alpha an | vedclass

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Question

(C) Given the polynomial $p(x) = 3x^2 - x - 4$,comparing it with $ax^2 + bx + c$,we have $a = 3$,$b = -1$,and $c = -4$.The sum of the zeros is $\alpha

Vedclass · Accepted Answer

$-\frac{4}{9}$

Vedclass · Answer

(C) Given the polynomial $p(x) = 3x^2 - x - 4$,comparing it with $ax^2 + bx + c$,we have $a = 3$,$b = -1$,and $c = -4$.The sum of the zeros is $\alpha + \beta = -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}$.The product of the zeros is $\alpha \beta = \frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$.We need to find the value of $\alpha^2 \beta + \alpha \beta^2$.Factoring out $\alpha \beta$,we get $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta(\alpha + \beta)$.Substituting the values,we get $\left(-\frac{4}{3}\right) \left(\frac{1}{3}\right) = -\frac{4}{9}$.

The zeros of $p(x) = 3x^2 - x - 4$ are $\alpha$ and $\beta$,then $\alpha^2 \beta + \alpha \beta^2 = \ldots$

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