Question

find a cubic polynomial with the sum Sum of the product of its zeros taken two at a time and product of zeroes are 3, -1 and -3 respectively


This is my 4th time asking the same question everyone is giving wrong answer .please give me right answer please any maths expert


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Answer 1

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

• ✦ sum of zeroes = 3
• ✦ sum of the product of zeroes = -1
• ✦ product of zeroes = -3

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

• ✦we need to find the cubic polynomial.

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

Let α ,β ,γ be the zeroes of the required polynomial.

Then,

• α + β + γ = 3
• αβ + βγ + αγ = -1
• αβγ = -3

So,

The required polynomial is :-

$\small \red{ \rm\:p(x) = {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma} \: \\ \\ \longmapsto \rm \: {x}^{3} - ( 3) {x}^{2} + ( - 1)x - ( -3) \\ \\ \longmapsto \bf \: {x}^{ 3} - 3 {x}^{2} - x + 3 = 0$

So ,

The cubic polynomial is$\bf\pink{{x}^{ 3} -3 {x}^{2} - x + 3 = 0}$

Verification:-

p(x) = x³ - 3x² - x + 3 = 0

• a = 1
• b = -3
• c = -1
• d = 3

sum of zeroes = - b/a

α + β + γ = - b/a

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

sum of product of zeroes = c/a

αβ + βγ + αγ = c/a

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

product of zeroes = - d/a

αβγ = - d/a

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

LHS = RHS

hence Verified .

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Answer 2

Given.

• α + β + γ = 3
• αβ + βγ + αγ = -1
• αβγ = -3

As we know that , the cubic polynomial is given by

$\star \: \: \sf \fbox{{(x)}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \alpha \gamma )x - \alpha \beta \gamma}$

Thus ,

$\sf \Rightarrow {(x)}^{2} - 3 {(x)}^{2} + (-1)x - (-3) \\ \\ \Rightarrow \sf {(x)}^{2} - 3 {(x)}^{2} -1x + 3$

$\therefore \sf \bold{ \underline{The \: cubic \: polynomial \: is \: {(x)}^{2} - 3 {(x)}^{2} -1x +3 }}$