Question

find the polynomial equation whose roots are 1,-1,3​

Answer 1

Answer:It is given that the roots are −2,1,3Hence the equation is (x+2)(x−1)(x−3)=0(x 2 +x−2)(x−3)=0x 3 +x 2 −2x−3x 2 −3x+6=0x 3 −2x 2 −5x+6=0This is the required cubic equation.please mark as brainlist

Answer 2

The polynomial equation whose roots are 1 , - 1 , 3 is

$\displaystyle \bf{ {x}^{3} - 3 {x}^{2} - x + 3 = 0 }$

Given :

The roots 1 , - 1 , 3

To find :

The polynomial equation whose roots are 1 , - 1 , 3

Solution :

Step 1 of 2 :

Write down the given roots

Here the given roots of the polynomial equation are 1 , - 1 , 3

Step 2 of 2 :

Find the polynomial equation

We know that if a , b , c are roots are of a polynomial equation then the polynomial equation is

(x - a) (x - b) (x - c) = 0

Hence the required polynomial equation is given by

$\displaystyle \sf{(x - 1)(x + 1)(x - 3) = 0 }$

$\displaystyle \sf{ \implies ( {x}^{2} - {1}^{2})(x - 3) = 0 }$

$\displaystyle \sf{ \implies ( {x}^{2} - 1)(x - 3) = 0 }$

$\displaystyle \sf{ \implies {x}^{2}(x - 3) - 1(x - 3) = 0 }$

$\displaystyle \sf{ \implies {x}^{3} - 3 {x}^{2} - x + 3 = 0 }$

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