Question

(a) -3
22. If a, b, y be the zeros of the polynomial p(x) such that (a + B + 7) = 3,
(aß + By + ya) = -10 and aby = -24 then p(x) = ?​

Answer 1

$\large{\underline{\bf{\purple{Given:-}}}}$

• ✦ sum of zeroes = 3
• ✦ sum of product of zeroes = -10
• ✦ product of zeroes = -24

$\:$

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

• ✦ we need to find the polynomial

$\huge{\underline{\bf{\red{Solution:-}}}}$

$\:$

If α , β, and γ are the zeroes of the polynomial.

Then,

$\:$

[ x³- ( α + β + γ)x² +(αβ +βγ +γα )x -αβγ ]

$\:$

• α + β + γ = 3
• αβ + βγ + αγ = -10
• αβγ = -24

p(x) = [ x³- ( 3)x² + (-10 )x - (-24) ]

p(x) = x³- 3x² - 10x + 24 = 0

$\:$

So, the required cubic polynomial is

p(x) = x³- 3x² - 10x + 24 = 0

$\:$

Verification :-

• a = 1
• b = -3
• c = -10
• d = 24

$\:$

sum of zeroes = - b/a

α + β + γ = - b/a

$\\ \longmapsto \rm\:3 = \frac{-(-3)}{1}\:\\\longmapsto \rm\: 3=3$

αβ + βγ + αγ = c/a

$\\\longmapsto \rm\: -10=\frac{-10}{1}\:\\\longmapsto \rm\: -10= -10$

αβγ = - d/a

$\longmapsto \rm\: -24=\frac{-24}{1}\:\\\longmapsto \rm\: -24 = -24$

LHS = RHS

$\:$

hence verified.

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Answer 2

$\huge \underline \red{answer}$

If α , β, and γ are the zeroes of the polynomial.

Then,

[ x³- ( α + β + γ)x² +(αβ +βγ +γα )x -αβγ ]

α + β + γ = 3

αβ + βγ + αγ = -10

αβγ = -24

p(x) = [ x³- ( 3)x² + (-10 )x - (-24) ]

p(x) = x³- 3x² - 10x + 24 = 0

So, the required cubic polynomial is

p(x) = x³- 3x² - 10x + 24 = 0

THANK YOU.