Factoring Polynomials and Solving Higher Degree Equations

Recall. With respect to division polynomials behave a lot like natural numbers. It is not always
possible to divide two polynomials and get a polynomial as a result. The result may sometimes be
a polynomial but in general we will get a rational expression . The best we can hope in general is
to get a quotient and a reminder. Last time we show that given two polynomials a(x) and b(x)
we can perform long division find a quotient polynomial r(x) and a remainder polynomial r(x) so
that:

where the degree of the remainder r(x) is less than the degree of the denominator b (x).
Sometimes it is more convenient to write Formula (1) as

a(x) = q(x)b(x) + r(x) (2)

Example 1. Find the quotient and the remainder of the division:

Answer. We will use the long division algorithm:

Therefore the quotient is x² − 3x − 9 and the remainder is −14. We can write this as

x³ − 5x² − 3x + 4 = (x² − 3x − 9)(x − 2) − 14

Let p(x) = x³ − 5x² − 3x + 4. Then the previous example tells us that

p(x) = q(x)(x − 2) − 14

where q(x) = x²−3x−9 is the quotient of the division. If we now want to evaluate the polynomial
p(x) for x = 2 we get

p(2) = q(2)(2 − 2) − 14
= q(2) · 0 − 14
= −14

So the result of the evaluation p(2) is the same as the remainder of the division of p(x) by (x − 2).
Notice that in the previous calculation it doesn’t really matter what q(2) is. No matter what q(2)
is it is multiplied by 0 so it is “killed”. Therefore, for any polynomial p(x) we have that p(2) is the
same as the quotient of the division p(x) ÷ (x − 2).

Example 2. What is the remainder of the division (x³ − 2x³ + 7x² − 5x + 4) ÷ (x − 1).

Answer. Let p(x) = x³ − 2x³ + 7x² − 5x + 4. As sume that the quotient is q(x) and r is the
remainder. We will have

p(x) = q(x)(x − 1) + r

Plugging x = 1 in the above equation gives :

p(1) = r

So the remainder is the result of evaluating p(x) at x = 1. Thus:

r = p(1)
= 1³ − 2 · 1³ + 7 · 1² − 5 · 1 + 4
= 1 − 2 + 7 − 5 + 4
= 5

Example 3. Let p(x) = x³ + 8. Find the remainder of the division:



Answer. If q(x) is the quotient and r the remainder then we will have:

p(x) = q(x)(x + 2) + r

Plugging x = −2 in the equation above gives: p(−2) = r. So the remainder is:

r = p(−2)
= (−2)³ + 8
= −8 + 8
= 0

Now let’s practice this ideas:

1. Find the remainder of the division

2. Find the remainder of the division: (x⁵ + x⁴ + 2x³ + 2x² − 2x + 7) ÷ (x + 1)

3. Find the remainder of the division:

We have then the following:

Fact 1. For any polynomial p(x) and any real number a , the remainder of the division p(x)÷ (x−a)
is always equal to p(a).

When the remainder of the division p(x) ÷ b(x) is 0, we say that b(x) divides p(x) or that b(x)
is a factor of p (x). We conclude this section by stating the following special case of Fact 1.

Fact 2. For a real number a and a polynomial p(x), x−a is a factor of p(x) exactly when p(a) = 0.

Factoring polynomials

In the previous chapter we show how we can get the simplified expanded form of a polynomial given
as a product of two (or more) polynomials. In this chapter we will examine the reverse problem,
we will start from the expanded form of the polynomial and we will try to write it as the product
of two or more polynomials. We will start with some termino logy :

A polynomial is called reducible if it has a non constant factor. In other words, a polynomial
p(x) is reducible if we can write it as a product of two polynomials

p(x) = a(x)b(x)

and neither a(x) nor b(x) are constants.

Example 4. The following polynomials are reducible:

A. x³ − 2x
B. x² − 4
C. x³ + 8
D. x² − 2x − 15
E. 6x⁴ + x³ − 22x² − 11x + 6

Justification. We show how each of these polynomials can be written as a product of other polynomials.

A x³ − 2x = x(x² − 2)

B x² − 4 = (x − 2)(x + 2)

C x³ + 8 = (x + 2)(x² − 2x + 4)

D x² − 2x − 15 = (x + 3)(x − 5)

E 6x⁴ + x³ − 22x² − 11x + 6 = (x + 1)(2x + 3)(x − 2)(3x − 1)

Later on we will see how we can get these results.

If a polynomial is not reducible it’s called irreducible. So irreducible polynomials have only
constant factors.

Recall from the previous section that if a polynomial p(x) has a factor of the form x−a, where
a is a real number then p(a) = 0. Therefore if no matter what value we substitute for x, p(x) never
evaluates to 0 we can conclude that p(x) has no factors of the form x−a. We see then that values
of x that make the polynomial evaluate to 0 are important so we give them a special name:

A root of a polynomial p(x) is a number a such that p(a) = 0.