Question

What are integral roots?

Answer 1

Hint: We first describe the concept of integral roots. Integral roots mean the roots of a polynomial being integer. We take an arbitrary polynomial and find the integral roots of the polynomial.

Complete step-by-step answer:
Integral roots mean the roots of a polynomial being integer. We take an example of a polynomial $ {{x}^{2}}+6x-72 $ to find its roots.
In case of $ {{x}^{2}}+6x-72 $ , we break the middle term $ 6x $ into two parts of $ 12x $ and $ -6x $ .
So, $ {{x}^{2}}+6x-72={{x}^{2}}+12x-6x-72 $ . We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives $ -72{{x}^{2}} $ . The grouping will be done for $ {{x}^{2}}+12x $ and $ -6x-72 $ .
We try to take the common numbers out.
For $ {{x}^{2}}+12x $ , we take $ x $ and get $ x\left( x+12 \right) $ .
For $ -6x-72 $ , we take $ -6 $ and get $ -6\left( x+12 \right) $ .
The equation becomes $ {{x}^{2}}+6x-72={{x}^{2}}+12x-6x-72=x\left( x+12 \right)-6\left( x+12 \right) $ .
Both the terms have $ \left( x+12 \right) $ in common. We take that term again and get
 $
   {{x}^{2}}+6x-72 \\
  =x\left( x+12 \right)-6\left( x+12 \right) \\
  =\left( x+12 \right)\left( x-6 \right) \\
$
Therefore, $ \left( x+12 \right)\left( x-6 \right)=0 $ has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
So, values of x are $ x=6,-12 $ .
Here $ x=6,-12 $ are integral roots of $ {{x}^{2}}+6x-72 $ .

Note: The numbers which satisfy the value of a polynomial are called its roots. The roots which are integers i.e., not irrational or imaginary are called integral roots. The numbers which satisfy the value of a polynomial are called its roots.