Question

find a cubic polynomial with the sum of zeros,sum of the product of its zeroes taken two at a time and the products of its zeroes are 3,2,-4,respectively​

Answer 1

Answer:

$\large{\underline{\boxed{\sf x^3 - 3x^2 +2x +4}}}$

Step-by-step explanation:

Let the zeroes of the cubic polynomial be α, β and γ respectively.

Given that -

Sum of zeroes = 3

⇒ α + β + γ = 3

Sum of the product of zeroes taken two at a time = 2

⇒ αβ + βγ + γα = 2

Product of zeroes = - 4

⇒ αβγ = - 4

Now, we know that ;

The cubic polynomial is given by ;

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

⇒ x³ - 3x² + 2x + 4

Hence, the required cubic polynomial is x³ - 3x² + 2x + 4.

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$\large{\underline{\underline{\mathfrak{\heartsuit \: Extra \: Information: \: \heartsuit}}}}$

For a cubic polynomial ax³ + bx² + cx + d, the zeroes are α, β and γ, where

• α + β + γ = -b/a
• αβ + βγ + γα = c/a
• αβγ = - d/a

Answer 2

$Cubic\ polynomial\ is\ of\ the\ form\ ax^3\pm bx^2\pm cx\pm d=0$

Given

$sum\ of\ zeroes=\alpha+\beta+\gamma\\ product\ of\ zeroes\ taken\ two\ at\ a\ time=\alpha \times \beta+ \beta \gamma+ \gamma \alpha\\ product\ of\ zeroes= \alpha \beta \gamma$

But, its given that

sum of zeroes is 3, product of zeroes taken two at a time is 2 and product of zeroes is -4.

$\therefore\; \alpha+ \beta+ \gamma=3\\ \alpha \beta+ \beta \gamma+ \alpha\gamma=2$