Class 10 Math All topics Polynomials

Let $P(x)$ be a cubic polynomial with zeroes $α,β,γ$ if $P(0)P(21 )+P(−21 ) =100$find $αβ1 +βγ1 +γa1 $.

(a)

14

Correct answer: (a)

Solution: Let $P(x)=ax_{3}+bx_{2}+cx+d$

So $P(21 )+P(−21 )=2b(41 )+2d$

and $P(0)=d$

$⇒d24b +2d =100$

$⇒4d2b +2=100⇒db =196$

also $α+β+γ=−ab ,αβγ=−ad $

Hence $αβ1 +βγ1 +γa1 =αβγα+β+γ =196 =14$

So $P(21 )+P(−21 )=2b(41 )+2d$

and $P(0)=d$

$⇒d24b +2d =100$

$⇒4d2b +2=100⇒db =196$

also $α+β+γ=−ab ,αβγ=−ad $

Hence $αβ1 +βγ1 +γa1 =αβγα+β+γ =196 =14$

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introduction to trigonometry

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introduction to trigonometry

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