Question

find the polynomial p(x) whose zeroes are -1 and 2​

Answer 1

Answer:

$\alpha = - 1 \\ \beta = 2 \\ \alpha + \beta = - 1 + 2 = 1 \\ \alpha \beta = - 1 + \times 2 = - 2 \\ p(x) = {x}^{2} - x - 2$

Answer 2

Here, the zeroes are -1 and 2.

We know that the sum of the zeroes =

$\large \blue {\tt {\alpha + \beta = - 1 + 2 = 1}}$

And the product of the zeroes =

$\large \pink {\tt { \alpha \times \beta = ( - 1)(2) = - 2}}$

So, let the polynomial p(x) be

$\large \red {\tt {{x}^{2} + ( \alpha + \beta )x + ( \alpha )( \beta )}}$

So, according to the question,

$\large \green {\tt {{x}^{2} + ( \alpha + \beta )x + ( \alpha )( \beta )}}$

By using the sum of the zeroes and the product of the zeroes, we get

$\large \orange {\tt { {x}^{2} + (1)x + ( - 2) = {x}^{2} + x - 2}}$

Answer:

The polynomial is $\huge \purple {\tt {{x}^{2} + x - 2}}$