Download Polynomial Factoring: Degree, Leading Term, Rational Zeros, Synthetic Division and more Study notes Pre-Calculus in PDF only on Docsity!
Factoring a Polynomial
Vincent Newberry
De nitions
- The degree of a polynomial f (x) is the largest power of x in the polynomial.
- The leading term of a polynomial is the term with the highest power of x. The term that determines the degree of they polynomial.
- The leading coe cient of a polynomial is the coe cient of the leading term.
Example
- The degree of f (x) = 4x^5 + 3x^2 − 1 is 5.
- The leading term of f (x) = 4x^5 + 3x^2 − 1 is 4 x^5.
- The leading coe cient of f (x) = 4x^5 + 3x^2 − 1 is 4.
Rational Zero's Theorem
Theorem. The set composed of every factor of the constant term of a polynomial f (x) divided by every factor of its leading coe cient is the set of all possible rational roots of f (x).
Using Synthetic Division to Factor Polynomials
Steps
- Use the Rational Zeros Theorem to make the list of all possible rational roots.
- Test possible roots using synthetic division. Once you nd a root, rewrite the original polynomial with the root you just found factored out using the resulting coe cients from the successful synthetic division.
- Keep testing roots using the new, reduced coe cients and continuing to factor the polynomial until it is factored entirely into linear factors.
Example
Lets factor the polynomial f (x) = 4x^4 − 8 x^3 − 3 x^2 + 7x − 2.
- First we compile the list of all possible rational roots using the Rational Zero' Theorem. The factors of the constant term, 2, are ± 1 and ± 2. The factors of the leading coe cient, 4, ± 1 , ± 2 , and ± 4. So now we divide all the factors of{ 2 by all factors of 4 to get the following list: ± 1 , ± 2 , ±
- Now we start testing values until we nd a root. We have no choice but to choose at random.
4 − 8 − 3 7 − 2 1 4 − 4 − 7 0 4 − 4 − 7 0 − 2
This was not a root so we try another.
4 − 8 − 3 7 − 2 − 1 − 4 12 − 9 2 4 − 12 9 − 2 0
This time we found a root. This division tells us that we can factor f (x) as follows.
f (x) = (x + 1)(4x^3 − 12 x^2 + 9x − 2) Now we continue testing numbers with synthetic division to nd more roots. How- ever, now we try to divide 4 x^3 − 12 x^2 + 9x − 2.
4 − 12 9 − 2 − 2 − 8 40 − 98 4 − 20 49 − 100
Not a root so we try another.
4 − 12 9 − 2 2 8 − 8 2 4 − 4 1 0
We found another root! So that means we can factor
4 x^3 − 12 x^2 + 9x − 2 into (x − 2)(4x^2 − 4 x + 1)
We can factor 4 x^2 − 4 x + 1 by hand into (2x − 1)(2x − 1). So our entire polynomial f (x) factors the following way:
4 x^4 − 8 x^3 − 3 x^2 + 7x − 2 = (x + 1)(4x^3 − 12 x^2 + 9x − 2) = (x + 1)(x − 2)(4x^2 − 4 x + 1) = (x + 1)(x − 2)(2x − 1)(2x − 1)
