Question

P(x)=x^2+x - 6 find the zeroes of the polynomial

Answer 1

${\boxed{ \Huge{ \red{ \rm{ \underline{ {x}^{2} \: + \: x \: - \: 6}}}}}}$

$\bf \Large \implies \: \: {x}^{2} \: + \: x \: - \: 6 \\ \\ \bf \Large \implies \: \: {x}^{2} \: + \: 3x \: - \: 2x \: - 6 \\ \\ \bf \Large \implies \: \: x \: (x + 3) \: - 2 \: (x - 3) \\ \\ \bf \Large \implies \: \: (x + 3) \: \: (x - 2) \\ \\ \bf \Large \implies \: \: x + 3 = 0 \\ \\ \bf \Large \implies \: \: x = \: - 3 \\ \\ \bf \Large \implies \: \: x - 2 = 0 \\ \\ \bf \Large \implies \: \: x \: = \: 2 \\ \\ \\ \bf \Large \implies \: \: \: x \: \: = \: - 3 \: \: and \: \: 2$

${\boxed{ \large{ \blue{ \rm{ \underline{ \alpha \: = - 3}}}}}} \\ \\ {\boxed{ \large{ \blue{ \mathcal{ \underline{ \beta \: = \: 2}}}}}}$

Zeros of the polynomial is -3 and 2

Answer 2

$\fbox{\huge\pink{A}\blue{N}\purple{S}\green{W}\red{E}\orange{R}}$

QUESTION :- Find the zeroes of the polynomial given p(x) = x² + x - 6.

ANSWER :-

We have factorization the equation to find the values of alpha and beta.

${x}^{2} + x - 6$

$= {x}^{2} + 3x - 2x - 6$

$= x(x + 3) - 2(x + 3)$

$= (x + 3)(x - 2)$

Now we have factorized the equation.

The two values we have got are :-

$x + 3 = 0$ and $x - 2 = 0$

So values of are :-

$x = - 3$ and $x = 2$

$\fbox{\alpha = - 3}$ and $\fbox{\beta = 2}$

Zeroes of the polynomial are -3 and 2.