The example problem that we will be solving is the polynomial:
f(x)=x⁴-15x²+10x+24=0. In order to solve this problem we must first find the divisors of 24, which is the constant term. After doing so, we will then use the process called, Synthetic Division, to see which of these divisors will give us a remainder of zero. The divisor(s), that causes the remainders to be zero(0), will be the x's that are considered the solution(s) to this 4th degree polynomial equation. .
The divisors of 24 are: -1,1,-2,2,-3,3,-4,4,-6,6,-8,8,
-12,12,-24, and 24. We will now write down horizontally, from left to right, the coefficients of each term of the polynomial starting with the leading coefficient and ending with the constant term. We should put these numbers in Synthetic Division form. The following set of numbers, is the coefficients of the 4th degree polynomial:
1,0,-15, 10, and 24. Please note that the third-degree term, was missing in the polynomial equation, we still had to account for its coefficient which was zero(0).
The algorithm of Synthetic Division is; in the case of this example, we take the first coefficient,1 and multiply it by the first divisor, 1, which gives us the product, 1. We now add this product to the second coefficient 0, which gives us the sum 1, we then multiply this sum, 1, by the first divisor 1 and add it to the third coefficient, -15, and get the sum, -14. We continue this process, by repeating the steps that we have just done. That is; we multiply the sum, -14, by the first divisor, 1, which gives us the product, -14. We now add this product to the fourth coefficient 10, and get the sum, -4. We continue the process, by repeating the steps that we have just done. That is; we multiply the sum,
-4, by the first divisor, 1, which gives us the product,
-4, we now add this product to the fifth/last-term, the constant term, 24, and get the sum/remainder 20. since 20, is not equal to zero(0), then the divisor, 1, is NOT a root/solution to the given polynomial equation, if the last sum/remainder was zero(0), for the divisor 1, then x = 1, would have been a root/solution.
We now should, by trial and error, try the remaining divisors. Let us try the next divisor, -1. By applying the same process and steps as we did in Step(#3), we should see that -1 causes the last sum/remainder, to be zero( 0 ), hence -1 is a solution to this polynomial equation, and we can say x = -1, is a root of the equation.
We continue with our trial and error process. Since we have a solution, we will shorten our set of numbers, to be, the set of sums/remainders, that is, the new set of 'coefficients' are;
1,-1,-14,24. We now will try all the divisors including
-1 again, but excluding 1, since we can have repeated roots, but once a divisor, does not satisfy as root, it will never work again as root of the equation.
by trial and error we can try -1 again, with the new 'coefficients', 1,-1,-14,24, and we should see that,
-1, does not gives us a final sum/remainder of zero(0).
So since -1 is not a repeated root, we should move on to the other divisors, -2,2,-3,3, etc. We will see, as we try the other divisors, that 2, 3, and -4, would be the only other divisors that are roots of this polynomial.