find a cubic polynomial whose zeros …
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Sia ? 6 years, 6 months ago
Suppose {tex}\text{α,β and γ}{/tex} are the zeros of the said polynomial p(x)
Then, we have {tex}\alpha{/tex} = 2, {tex}\beta{/tex} = -3 and {tex}\gamma{/tex} = 4
Now,
{tex}\text{α+β +γ=2-3+4=3}{/tex}......(1)
{tex}\text{αβ +βγ+γα=2(-3)+(-3)(4)+(4)(2)=-6-12+8=-10......(.2)}{/tex}
{tex}\text{αβγ=2(-3)(4)=-24 .......(3)}{/tex}
Now, a cubic polynomial whose zeros are {tex}\alpha, \space \beta \space and \space \gamma{/tex} is given by
p(x) = x3 - {tex}\text{(α+β+γ)x}^2{/tex} + {tex}\text{(αβ+βγ+γα)x}{/tex} - {tex}\alpha\beta\gamma{/tex}
Now putting the values from (1),(2) and (3) we get
p(x) = x3 -(3)x2 + (-10)x - (-24)
= x3 -3x2 -10x +24
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