# Numerical methods for roots of polynomials 1

In short, the inner workings of the function are more exposed with this notation. Factoring Large Polynomials. Large polynomials larger than quadratics, equations involving powers of x larger than x 2 get harder to factor the bigger they get. While there are advanced techniques to directly calculate the roots of a cubic x 3 and in some cases a quartic x 4 , these methods are quite complicated and require an advanced sophistication in algebra to be comprehensible.

The reader is welcome to take a look at both of these cases to verify our opinion. We will concentrate on some theorems that offer factoring help on a less advanced basis. To be sure, use of these theorems amounts to educated guessing, but such guessing is actually more likely to get an answer faster than the advanced solution techniques.

At the least, the techniques we offer will show whether an elementary answer like an integer or a rational number can be expected. Failing that, we will explore a scheme for finding an answer numerically a refinement of trial and error using a calculator or computer. If the numerical technique is done carefully, we can sometimes use the decimal expansion calculated to guess a familiar irrational number.

If all else fails, we can resort to the "big guns" and use one of the advanced techniques. Fundamental Theorem of Algebra. The roots may be repeated i. Rational Zeros Theorem. Subtracting a 0 q n from both sides gives. Since each term on the left contains at least one p, we can factor it out:. The term in parentheses on the left is the sum of many products of integers and so is an integer. Call this integer I and we have. Since p and q have no common divisors other than 1, the same must be true of p and q n , which leaves p dividing a 0. We could have subtracted a n p n from the equation after multiplying through by q n , giving us.

From here the proof is similar, and is left as an exercise for the reader. The result is that q is proved to divide a n. Use the rational zeros theorem to guess the possible rational roots of. Then use synthetic division to find which of the possible roots is actually a root. According to the theorem, we are looking for numerators that divide 2 1 and 2 and denominators that divide 3 1 and 3.

Thus, the possible roots are:. We should point out that even if the seven wrong answers had to be tried first, synthetic division is fast enough to go through all the potential roots in a matter of minutes. Irreducible Expressions. It is the polynomial analogue of a prime number. If we allow complex number solutions, then the above equation has the solutions.

We will normally stop factoring a polynomial when we encounter an irreducible quotient, since exceptions are normally reserved for more advanced subject matter than that we cover here. For readers who can use a programmable calculator, the following method can be used to get a decimal expansion of a root. If one know the decimal expansions of a few irrational numbers, guesses can be made based on the computed decimal expansion and checked.

This method is a bit more cumbersome than the above guessing scheme for rational roots and should be tried only if one fails to find a rational root. The reader is advised that algebra and arithmetic errors are extremely common when learning to handle polynomials, and should be eliminated first before trying a numeric solution.

## numerical methods - Polynomial root finding - Mathematics Stack Exchange

One way is to use the solve function. Another way is to use the factor function to factor the polynomial terms. A modified version of this example exists on your system. Do you want to open this version instead? Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select:.

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