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Factoring Polynomials and Solving Higher Degree Equations

Example 13. Factor 5x2(3x − 7) + 3x − 7.

Answer. We can think of the last two terms together as forming one summand :

5x2(3x − 7) + (3x − 7)

Once we insert the “missing” parentheses we can see that we have 3x − 7 as a common factor . So:

5x2(3x − 7) + 3x − 7 = (3x − 7)(5x2 + 1)

Factoring by grouping

Some times even though there is no common factor for all the terms of a polynomial, we can separate
the terms into two or more groups , in such a way that each of the groups has a common factor
and after we have factored these common factors the other factors are the same. Let’s see some

Example 14. Factor ax + ay + bx + by.

Answer. There is no common factor to all four terms. However the first two terms have common
factor a, and the last two terms have common factor b. After ”factoring” out these common factors
we have:

a(x + y) + b(x + y)

The two summands now have a common factor x + y. So we can factor it out:

(a + b)(x + y)

Example 15. Factor ax − ay + bx − by + cy − cx

Answer. Again there is no factor common to all the terms of this polynomial. However the first
two terms have a as a common factor, the middle two terms have b as a common factor, and the
last two terms have c as a common factor. After ”factoring out” these common factors we get:

a(x − y) + b(x − y) + c(y − x)

Each of the three summands now has x − y as a common factor. So we finally get:

(x − y)(a + b − c)

Alternatively, we could have noticed that the first, third and sixth term have x as a common
factor and the second, fourth and fifth have y as a common factor. Then we get:

x(a + b − c) − y(a + b − c)

which finally gives

(a + b − c)(x − y)

Notice that this is the same factoring, really . It has the same two factors but in different order .
This is typical, usually there will be more than one ways to group the terms of a polynomial so
that each group has a common factor and the after factoring the other terms are the same.

Example 16. Factor 10x2y3 + 8wxy − 15wxy2 − 12w2

Answer. We have:

10x2y3 + 8wxy−15wxy2−12w2 = 5xy2(2xy − 3w) + 4w(2xy − 3w)
= (2xy − 3w)(5xy2 + 4w)

Let’s practice this technique:

1. Factor: 5za2 + 3xa2 − 5bz − 3bx

2. Factor: 10yx2 − 8x2 + 15xy − 12x

3. Factor: 4yx2z3 + 10zx3 − 6y2z2 − 15xy

4. Factor: 2a2x2 − 5ya2 + 5by − 2bx2 + 2c3x2 − 5yc3

Factoring trinomials by splitting the linear term

In this section we concentrate on quadratic polynomials in one variable . Such a polynomial must
have a quadratic term, and it may (or may not) have a linear term and a constant term. It is
customary to use the letters a, b and c for the coefficients of the quadratic , the linear and the
constant term respectively. So the form of the polynomials we will deal with in this section is:

p(x) = ax2 + bx + c, a ≠ 0

The quadratic term has to be present , hence its coefficient a cannot be 0. The linear term and the
constant term however may be missing, that is b and/or c may be 0.

For the remaining of this section, a, b, and c will stand for the coefficients
of the quadratic, the linear, and the constant term respectively.

Example 17. Identify a, b, and c for each of the fol lowing polynomials :

A. 2x2 − x + 6
B. −2x2 − 5
C. x2 + 5x
D. 7x2


A. a = 2, b = −1, c = 6

B. a = −2, b = 0, c = −5

C. a = 1, b = 5, c = 0

D. a = 7, b = 0, c = 0

The method we will use consists of “splitting” the linear term into the sum of two terms in
such a way that grouping one of these new terms with the quadratic term and the other with the
constant term “works”. Before explaining how to choose this splitting let’s see a couple of examples:

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