Now I will explain how to apply Gauss’ theorem to factor a polynomial into factors. This theorem must be combined with the theorem of the rest and with the properties of the roots to obtain the decomposition of the polynomial.
Gauss’ theorem tells us that the possible roots of a polynomial are obtained by means of the quotient between the divisors of the independent term and the divisors of the main coefficient (coefficient of the term with the highest degree).
For example, let’s imagine we have a grade 4 polynomial:
Its possible roots would be all the ratios of each coefficient divisor e and each coefficient divisor a:
We would have to make the quotients of all the combinations of each of the divisors of the independent term between each divisor of the main coefficient.
To calculate which of these possible roots correspond to the roots of the polynomial, we apply the theorem of the rest, that is to say, the roots will be those that make the value of the polynomial zero.
Once the roots are obtained, the polynomial can be expressed as:
Where a is the main coefficient and the different x1, x2, x3… are the roots of the polynomial. I remind you that the number of roots of a polynomial coincides with the degree of that polynomial.
Gauss’s theorem exercise to break down polynomials
Let’s look at all this with an example for you to assimilate better: apply Gauss’ theorem to break down the following polynomial:
First, we are going to simplify all the terms of the polynomial by dividing them by 3 to simplify the whole procedure:
We do this because having a main coefficient of 1 as the main coefficient, the possible roots will coincide with the divisors of the independent term and we will not have to combine both divisors to find the roots. In this way, the number of possible roots is considerably reduced.
The prime divisors of the independent term, which in this case is 24 are:
But let’s not forget, we also have compound dividers, as they are:
To check which of these possible roots correspond to the roots of the polynomial, we have no choice but to apply the theorem of the rest, checking which of them makes the value of the polynomial 0.
As the polynomial is of degree 4, we must find 4 roots among all the possible ones.
We start with the smaller dividers, as the operations are simpler. We start by checking the value of the polynomial with 1:
The value of the polynomial is equal to 0, so 1 is the root of the polynomial:
We’re still down to -1:
The value of the polynomial is other than zero, so -1 is not a root of the polynomial:
We’re still on two:
The value of the polynomial is equal to -24, other than zero, so 2 is not a root of the polynomial:
We’re still down to -2:
The value of the polynomial with x=-2 is equal to 0, so -2 is the second root of the polynomial.
We have two more roots left.
We tried 3:
The value of the polynomial is 0, so x=3 is another root of the polynomial:
We’ve already found three roots. We only have one more left.
We’re still trying x=-3:
The value of the polynomial with x=-3 is not equal to 0, so -3 is not the root of the polynomial either:
We’re still down to four:
The value of the polynomial is not 0 with x=4, so 4 is not the root of the polynomial:
We tried x=-4:
The value of the polynomial xon xon x=-4 is 0. Therefore x=-4 is another root of the polynomial:
And with this root we already have the four roots we were looking for, which are these:
According to Gauss’ method of factoring polynomials, the decomposed polynomial has this form:
So we have to replace the value of a and its roots with its values:
Finally we operate and we’re left with it:
As you can see, decomposing polynomials using Gauss’ theorem is to test them with different possible roots until you find them one by one, so it is not a direct method of decomposing polynomials.
It is even less direct than the Ruffini method, since with the Ruffini method, as we move forward, we discard more possibilities.