• Theory and Examples
• Exercises

### Why should we rewrite terms as products?

Factoring a term means that this term is transformed into a product of two or more factors.

Why should we do this? Products are important to

• reduce fractions and to
• solve equations.

In the following, we present an example for each case. The individual steps of the solution are not commented but they will be covered later in this chapter.

Example 1

Reduce the fraction
$\large \frac{{{a^2} - 2a}}{{{a^2} - a - 2}}$

Solving by factoring both the numerator and denominator:
$\large \frac{{{a^2} - 2a}}{{{a^2} - a - 2}} = \frac{{a \cdot (a - 2)}}{{(a + 1) \cdot (a - 2)}} = \frac{a}{{(a + 1)}}$

More details are available in the chapter Division and Fractions.

Example 2

Solve the equation

$2{x^3} - 4{x^2} + 2x = 0$

Solving by factoring the terms step by step on the left hand side:

$2x \cdot ({x^2} - 2x + 1) = 0$

$2x \cdot {(x - 1)^2} = 0$

The product equals zero if one of the two terms is zero. Therefore, the equation has two solutions:

${x_1} = 0 \qquad {x_2} = 1$

More details are available in the chapter Equations - Basics.

There are different factorization methods. You should be able to apply at least the following two methods.

### Method 1: Factoring out a common factor

A factor which appears in all summands can be placed in front of the bracket.

Example 1

$2a + 6 = 2 \cdot (a + 3)$

Factoring out the common factor 2

Example 2

$6{x^2} - 15xy = 3x \cdot (2x - 5y)$

Factoring out the factors 3 and x

Example 3

$2pq + q = q \cdot (2p + 1)$

Factoring out the common factor q
The number 1 inside the bracket replicates the second summand!

Technically, factoring out is an application of the distributive law:

$a \cdot (b + c) = a \cdot b + a \cdot c$

Remark on the notation:

The multiplication sign might be omitted. The factor which is factored out can also be placed either ahead or behind the bracket (commutative law of multiplication):

Example 4

${h^2} - 2h = h(h - 2)$

Example 5

$4xy + 10y = (2x + 5) \cdot 2y$

The original term can also consist of more than two summands:

Example 6

$15{b^2} - 10ab + 20b = 5b\left( {3b - 2a + 4} \right)$

In many cases it makes sense to factor out a combination of terms:

Example 7

$7a \cdot (x - 2) + 3b \cdot (x - 2) = (x - 2) \cdot (7a + 3b)$

In some cases, a full factorization can be obtained step by step several factorizations:

Example 8

$10ax - 15ay + 8bx - 12by = 5a(2x - 3y) + 4b(2x - 3y) = (5a + 4b)(2x - 3y)$

After factoring out 5a in the leading terms and 4b in the remaining ones, both summands contain the factor (2x - 3y). This term can be placed in front of the new bracket.

Of course, it is also possible to facor out x in the first and third elements, and y in the second and fourth one. The result would of course be the same.

Remark:

Factorization might be impossible in some cases.

### Method 2:  Factorization of quadratic terms

Example 1

A sum of several terms such as

${x^2} - 6x + 9$

cannot be factored out, since the three summands do not have a common factor.

Nonetheless, this term can be factored out because it represents the "right-hand side" of the second binomial formula:

${(x - 3)^2} = {x^2} - 6x + 9$,   so  ${x^2} - 6x + 9 = (x - 3) \cdot (x - 3)$

An application of the binomial formulas for a factorization can be found in the following two examples:

Example 2

$4{a^2} + 4ab + {b^2} = {(2a + b)^2}$ because we have  ${(a + b)^2} = {a^2} + 2ab + {b^2}$

Example 3

${x^2} - 25 = (x + 5)(x - 5)$          because we have  $(a + b)(a - b) = {a^2} - {b^2}$

Please note: In order to apply the binomial formulas two summands must consist of quadratic numbers (1, 4, 9, 16, 25, 36, 49, 64,...).
Approach: First, you should take the roots of the quadratic terms. Then you should control "backwards" by checking whether the mixed term is correct!

Even if there are no quadratic numbers, quadratic terms might be decomposed into binomial factors ("bracket times bracket"):

Example 4

${a^2} + 4a - 12 = (a + 6)(a - 2)$

Example 5

$3{w^2} + w - 2 = (w + 1)(3w - 2)$

However, even if this can be done, it is quite difficult to find the solution by "trial end error". If the quadratic summand in example 4 has 1 as its coefficient ($1 \cdot {a^2}$) there is a helpful rule:

Both numbers in the bracket must sum up to the number in the second element. Moreover, both numbers in the bracket multiplied by each other must result in the third element.

In example 5, this means:    sum $6 + ( - 2) = 4$,  product  $6 \cdot ( - 2) = - 12$

Hint: Start with the possible comibinations of numbers for the product and check whether these numbers add up to what they are supposed to!

Further examples:

Example 6

${m^2} - 3m + 2 = (m - 1)(m - 2)$

because:  $( - 1) + ( - 2) = - 3$,  $( - 1) \cdot ( - 2) = + 2$

Example 7

${a^2} + 2ab - 15{b^2} = (a + 5b)(a - 3b)$

because:  $5 + ( - 3) = 2$,  $5 \cdot ( - 3) = - 15$

### A combination of the two methods is possible

The two methods for factorization can be combined. We recommend to factor out common factors for quadratic terms first. Then, one should try to decompose the terms into binomials.

Example 1

$7{u^2} - 28{v^2} = 7({u^2} - 4{v^2}) = 7(u + 2v)(u - 2v)$

Example 2

$12{x^3} + 60{x^2} + 48x = 12x({x^2} + 5x + 4) = 12x(x + 1)(x + 4)$

1)

Factor out as much as possible:

a)

$3x + 3y$

b)

$8{t^2} - 4t$

c)

$15ab + 5ac - 25ad$

d)

$12{m^2}n + 20m{n^2}$

e)

$3a(x - 2) - (x - 2)$

2)

Decompose by multiple factorization:

a)

${a^2} + 2ab + ac + 2bc$

b)

$6y - 9 - 2xy + 3x$

3)

Factor out using the binomial formulas:

a)

${a^2} + 8ab + 16{b^2}$

b)

$9{v^2} - {w^2}$

c)

$4{r^2} - 4rs + {s^2}$

d)

${x^4} + 2{x^2} + 1$

4)

Decompose into two brackets:

a)

${x^2} + 4x - 5$

b)

${y^2} - y - 20$

c)

$3{p^2} - p - 2$

d)

${a^2} - 10ab + 16{b^2}$

5)

Use one or multiple methods for factorization in order to receive as many factors as possible:

a)

$2{c^2} - 18{d^2}$

b)

$3{x^2}y - 15xy + 12y$

c)

$2{w^3} - 20{w^2} + 50w$

d)

$3(t - 2) + ({t^2} - 4)$

e)

${p^4} - 81$

f)

$4a{x^2} - a(4x - 1)$