Factoring Polynomials

Engineering Jump Start - Summer 2012 – Mathematics Worksheet #3
Factoring Polynomials
Concepts: A linear combination of quantities is a sum of arbitrary multiples of these quantities.
For example, one linear combination of A, B, C and D is 2𝐴 + 3𝐡 βˆ’ 5𝐢 + 4𝐷.
A polynomial in x is a linear combination of non-negative integer powers of x. For example,
𝑓(π‘₯) = βˆ’2π‘₯ 5 + 3π‘₯ 3 + 2π‘₯ + 1 is a polynomial in x. The numbers in front of the powers of x are
also known as the coefficients. A real polynomial is a polynomial that has only real coefficients.
Check point:
Is 3π‘₯ 5 + π‘₯ 3 + π‘₯ + π‘₯ βˆ’1 a polynomial? Why or why not?
1
Is βˆ’π‘₯ 3 + 2π‘₯ + π‘₯ 2 a polynomial? Why or why not?
An important problems associated with polynomials is factoring. One reason why factoring is
important is that we know that a product is zero precisely when one or more of its factors are
zero. Thus, having a factorization allows us to find the zeros of a polynomial.
Another important application of factoring in calculus is to be able to find the so-called partial
fraction decomposition.
How to factor polynomials
If you know that a polynomial is divisible by a linear factor, then you can apply synthetic
division. But how do you come up with linear factors that divide the polynomial in the first
place?
Let’s consider the example of the polynomial π‘₯ 3 + 1. It’s possible to know instantly, without
attempting trial-and-error factorizations that π‘₯ + 1 must be a linear factor of that polynomial,
in other words, that π‘₯ 3 + 1 can be written as (π‘₯ + 1) times another polynomial.
Why? Try to recall the β€œfactor theorem” from precalculus.
Student Exercises:
1. Explain why π‘₯ 3 + 1 must be divisible by π‘₯ + 1 without performing long or synthetic
division.
Engineering Jump Start - Summer 2012 – Mathematics Worksheet #3
2. Use synthetic division to factor π‘₯ 3 + 1 into (π‘₯ + 1) times a quadratic polynomial.
3. This quadratic polynomial does not have real zeros. Complete the square to
demonstrate that, and then use that to find the complex zeros. (A real polynomial
without real zeros is called irreducible).
4. State the theorem you learned in precalculus about the imaginary zeros of real
polynomials.
5. Using the factor theorem, factor the quadratic polynomial you found in 2. into a product
of linear factors.
In the previous exercise, we factored a cubic polynomial with real coefficients into both a
real linear factor times an irreducible quadratic, and a product of linear factors, some of
them with imaginary numbers. The fundamental theorem of algebra guarantees that that is
always possible:
Fundamental theorem of algebra: any polynomial 𝑝(π‘₯) of degree n with complex
coefficients and leading coefficient π‘Žπ‘› can always be written as
𝑝(π‘₯) = π‘Žπ‘› (π‘₯ βˆ’ 𝑐1 )(π‘₯ βˆ’ 𝑐2 ) … (π‘₯ βˆ’ 𝑐𝑛 )
where 𝑐1, 𝑐2 … 𝑐𝑛 are complex numbers.
Student Exercises:
1. Demonstrate that if 𝑐1 = π‘Ž + 𝑏𝑖 and 𝑐2 = π‘Ž βˆ’ 𝑏𝑖, with real numbers a,b and b not zero,
then (π‘₯ βˆ’ 𝑐1 )(π‘₯ βˆ’ 𝑐2 ) is a real and irreducible quadratic.
2. The fundamental theorem only guarantees that polynomials can be completely
factorized into the leading coefficient times a product of linear factors. Why is it, then,
that real polynomials can be factored completely into real linear factors and irreducible
quadratics?
3. Factor the following polynomials completely into linear factors and irreducible
quadratics:
Engineering Jump Start - Summer 2012 – Mathematics Worksheet #3
a. 𝑝(π‘₯) = π‘₯ 4 βˆ’ 1
b. 𝑝(π‘₯) = π‘₯ 6 βˆ’ 1
c. 𝑝(π‘₯) = π‘₯ 5 βˆ’ 5π‘₯ 3 + 4π‘₯
d. 𝑝(π‘₯) = π‘₯ 4 + 2π‘₯ 2 + 1
e. (harder) 𝑝(π‘₯) = π‘₯ 8 βˆ’ 1