Question

write the polynomial in x whose zeroes are 1,2and -1

Answer 1

Given:

Zeroes of the polynomial are 1,2,-1

To find:

The polynomial

Explanation:

$\sf if \: 1,2,-1 \: are \: the \: zeroes \: then,$

$\sf (x-1),(x-2),(x+1) \: are \: the \: factors \: of \: polynomial$

Let the polynomial be f(x)

$\longrightarrow \: \sf p(x) = (x - 2)(x - 1)(x + 1)$

Using the formula

$\boxed{\sf (x+y)(x-y)=x^2-y^2}$

$\longrightarrow \: \sf p(x) = (x - 2)( {x}^{2} - 1) \\ \\ \longrightarrow \: \sf p(x) = {x}^{3} - x - 2 {x}^{2} + 2 \\ \\ \longrightarrow \: \boxed{ \sf p(x) = {x}^{3} - 2 {x}^{2} - x + 2 }$

Therefore,the polynomial is x³-2x²-x+2

Extra information:

If α, β ,δ are the roots of a cubic polynomial

then

➡α+β+δ=-b/a

➡αβ+βδ+αδ=c/a

➡αβδ=-d/a

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Let a,b and c be the zeros of required cubic polynomial

Given

a = 1 ,b = 2 and c = - 1

Sum of Zeros : - x² coefficient / x³ coefficient

a + b + c

= 1 + 2 + (-1)

→ a + b + c = 2................[1]

Product of Zeros : constant term / x³ coefficient

abc = 1(2)(-1)

→ abc = -2..........[2]

Product of successive zeros : x coefficient / x³ coefficient

ab + bc + ac

= (1)(2) + (2)(-1) + (-1)(1)

= 2 - 2 - 1

→ ab + bc + ac = -1...........[3]

Required Polynomial

x³ - (a + b + c)x² + (ab + bc + ca)x - abc

From equations [1],[2] and [3],we get:

→ x³ - 2x² - x + 2