Question

Write the zeroes of the polynomial -${{\text{x}}^2} - {\text{x}} - 6$
A) $4,7$ B) $2, - 4$ C) $3, - 2$ D) $1,7$

Answer 1

Hint: Zeroes of the polynomial are the real values of the variable for which the value of polynomial becomes zero. You can factorize the given polynomial and put the factors equal to zero. Then find the values of the variable x.

Complete step by step answer:

Given polynomial is-${{\text{x}}^2} - {\text{x}} - 6$. Now let f(x) =${{\text{x}}^2} - {\text{x}} - 6$
We have to find the zeroes of the given polynomial. Zeroes of the polynomial are the real values of the variable for which the value of polynomial becomes zero. So to find the zeroes of f(x), put f(x)$ = 0$.
$ \Rightarrow {{\text{x}}^2} - {\text{x}} - 6 = 0$
Now we will factorize the polynomial by splitting the middle term as we can write $ - {\text{x = - 3x + 2x}}$, so we get-
$ \Rightarrow {{\text{x}}^2} - 3{\text{x + 2x - 6 = 0}} \Rightarrow \left( {{{\text{x}}^2} - 3{\text{x}}} \right){\text{ + }}\left( {{\text{2x - 6}}} \right){\text{ = 0}}$
On taking x and $2$ common, we get-
$ \Rightarrow {\text{x}}\left( {{\text{x}} - 3} \right) + 2\left( {{\text{x - 3}}} \right) = 0$
By separating the common factor we get,
$
   \Rightarrow \left( {{\text{x - 3}}} \right) \times \left( {{\text{x + 2}}} \right) = 0 \\
   \Rightarrow \left( {{\text{x - 3}}} \right) = 0{\text{ or }}\left( {{\text{x + 2}}} \right) = 0 \\
 $
Now equating the values of x to zero, find the value of x,
$ \Rightarrow {\text{x}} = 3{\text{ or x = }} - 2$
Therefore, the zeroes of the polynomial ${{\text{x}}^2} - {\text{x}} - 6$ are $ - 2,3$ .
Hence, the answer is ‘C’.

Note: You can also use the formula ${\text{x = }}\dfrac{{ - {\text{b}} \pm \sqrt {{{\text{b}}^2} - 4{\text{ac}}} }}{{2{\text{a}}}}$ to find the zeroes of quadratic polynomial ${\text{a}}{{\text{x}}^2}{\text{ + bx + c}}$. This formula is known as discriminant formula. You just need to put the values and simplify. Here, b=$ - 1$, a=$1$ and c=$ - 6$. On putting these values in formula, we get-
$ \Rightarrow {\text{x}} = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1 \times \left( { - 6} \right)} }}{{2 \times 1}} = \dfrac{{1 \pm \sqrt {1 + 24} }}{2}$
On simplifying we get-
\[
   \Rightarrow {\text{x = }}\dfrac{{1 \pm \sqrt {25} }}{2} = \dfrac{{1 \pm 5}}{2} \\
   \Rightarrow {\text{x = }}\dfrac{{1 + 5}}{2}{\text{ or x = }}\dfrac{{1 - 5}}{2} \\
   \Rightarrow {\text{x = -2 or 3}} \\
 \]
Hence we get the same answer.