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Finding an integer solutions of a polynomial function

integer solution of polynomial functions

Integer solutions of a polynomial function theorem says: If a polynomial function a_n x^n + a_{n - 1} x^{n - 1} + ... + a_1 x + a_0 = 0 with integer coefficients has an integer solution, a \not= 0, then that solution is the divisor of free coefficient a_0 .

As an addition to this theorem, for every whole number k, number \alpha - k is a divisor of \ f(k).

integer solution of polynomial functions

Example 1. Find all integer roots of x^3 - 8x^2 + 25x - 26 = 0.

f(x) = x^3 - 8x^2 + 25x - 26

The set of dividers is, according to the last theorem, the set of dividers of number -26.

a \in \{-1, 1, -2, 2, -13, 13, -26, 26 \}

It would take a lot of time to check all these solutions so we’ll first eliminate few of them.

We can take any whole number k, for example here we’ll take k = 1. According to the addition to the theorem, for every integer root of this polynomial –\alpha , polynomial \alpha - 1 has to divide f(k) without the remainder.

Now let’s take a look at a set of solutions for \alpha - 1.

\alpha - 1 \in \{-2, 0, -3, 1, -14, 12, -27, 25 \}

f(1) = 1^3 - 8 * 1^2 + 25 * 1 - 26

f(1) = -8

Now we can eliminate every with which – 8 is not divisible.

\alpha - 1 \in {-2, 0, -3, 1, -14, 12, -27, 25}

\alpha \in \{-1, 2 \}

Now we have significantly smaller set of possible solutions. Now we have to check both of them to see are they roots of given equation.

integer-solution-of-polynomial-function
This means that number 2 is the only whole root of this equation.

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