Question

the sum of zeroes of a polynomial Kx³-3x²-2x+1 is 1 then Find value of K​

Answer 1

Step-by-step explanation:

$sum \: of \: the \: roots \: of \: cubic \: equation \: = - b \div a \\ - ( - 3) \div k = 1 \\ = > k = 1$

Answer 2

Answer:

HOW TO SOLVE?

$sum \: of \: zeros \: = - \frac{b}{a} \\ let \: the \: zeros \: be \: \alpha \: and \: \beta \: and \: \gamma \\ \alpha + \beta + \gamma = 1 \: \: \: \: \: (given) \\ \alpha + \beta + \gamma = \frac{ - ( - 3)}{k} \\ \alpha + \beta + \gamma = \frac{3}{k} = 1 \: \: \: \: (given) \\ \frac{3}{k} = 1 \\ k = 3$

The value of K is 3.

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