Question

If we are given roots as $\alpha ,\beta$ and 1 of the polynomial ${{x}^{3}}-2{{x}^{2}}-5x+6=0$, then find the value of $\alpha$ and $\beta$.

Answer 1

Hint: Here, 1 is a root so $x-1$ will be a factor. Then perform the polynomial division, reduce the polynomial ${{x}^{3}}-2{{x}^{2}}-5x+6=0$ from 3rd degree to 2nd degree polynomial to get the values for $\alpha$ and $\beta$.

Complete step-by-step solution -
Here, it is given that $\alpha ,\beta$ and 1 are the three roots of the polynomial ${{x}^{3}}-2{{x}^{2}}-5x+6=0$.
We have to find the values for $\alpha $ and $\beta $.
Since, 1 is a root we can say that $x-1$ is a factor of the polynomial ${{x}^{3}}-2{{x}^{2}}-5x+6=0$.
Next we have to reduce this third degree polynomial into a second degree polynomial by polynomial division.
Now, we can do the polynomial division:
$x-1\overset{{{x}^{2}}-x-6}{\overline{\left){\begin{align}
  & {{x}^{3}}-2{{x}^{2}}-5x+6 \\
 & {{x}^{3}}-{{x}^{2}} \\
 & \overline{0\text{ }-{{x}^{2}}\text{ }-5x} \\
 & \text{ }-{{x}^{2}}+\text{ }x \\
 & \text{ }\overline{0\text{ }-6x\text{ }+6} \\
 & \text{ }-6x\text{ }+6 \\
 & \text{ }\overline{\text{0 }+0} \\
\end{align}}\right.}}$
In polynomial division, like normal division after each step we have to do subtraction.
From the above polynomial division we will get:
$(x-1)({{x}^{2}}-x-6)=0$
Hence, from the above equation we can say that:
$x-1=0$ or ${{x}^{2}}-x-6=0$
Consider the equation ${{x}^{2}}-x-6=0$. It can be written as:
${{x}^{2}}-3x+2x-6=0\text{ }.....\text{ (1)}$
That is, $-x=-3x+2x$
From equation (1) $x$ is common for the first two terms and 2 is common for the last two terms.
Now, take outside $x$ from the first two terms and 2 from the last two terms, equation (1) becomes:
$x(x-3)+2(x-3)=0$
In the above equation $x-3$ is common, so take outside. Thus we obtain the equation:
$(x-3)(x+2)=0$
The above equation implies:
$x-3=0$ or $x+2=0$
$\begin{align}
  & x-3=0\Rightarrow x=3 \\
 & x+2=0\Rightarrow x=-2 \\
\end{align}$
Hence, we got the other two roots of the equation as $-2$ and $3$
That is, the values for $\alpha $ and $\beta $ are $\alpha =-2$ and $\beta =3$
Hence, the three roots of the equation are $1,-2$ and $3$

Note: Here, for reducing the polynomial from 3rd degree to 2nd degree instead of polynomial division you can do the synthetic division also. i.e. by taking the coefficients. For the quadratic equation you can also find the roots directly by putting the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.