# Polynomials

Polynomials

**Factoring**

The process of factoring is a type of ”reverse-multiplication”, where you are
given a polynomial and have to

write it as a product of factors. A polynomial is completely factored when each
factor is prime, or cannot be

factored again. The idea is very similar to factoring numbers: 60 =5
·12 is

one factoring of 60

is the prime factorization of 60

There are several strategies to determine how to factor polynomials.

**I. Common Factors **

A common factor is a factor of every term of an
expression . Common factors can be pulled out of an

expression using the distributive property in reverse :

**Examples:
**
(Note the negative in the second term)

Finding common factors should always be the first step in factoring an
expression.

**II. Factoring by Grouping**

Factoring by Grouping is especially useful when you have
more than three terms in the polynomial. The

technique of factoring by grouping is really using common factors creatively, as
shown in the following

example:

**Example 1: **Factor 2x^{3} - x^{2} + 6x - 3

Solution : Although there is no factor common to all terms (except 1, which we
ignore), we can ”group”

the polynomial by twos, each of which have a common factor:

Now notice that there are two terms , each of which has the factor 2x - 1.

It is very easy to check your answer by mulitplying it out:

**III. Factoring Using Special Products**

The special products we learned in the section ”Products of Polynomials” can be
used to help factor

expressions which have an appropriate form:

**Example 2:** Factor completely x^{3 }- 9x

Solution: As mentioned in the previous section, the first thing we do is look
for a common factor:

Notice how the second factor can now be written as x^{2}
- 3^{2} ,
a difference of two squares. Recall that

The expression is now completely factored.

**Example 3: **Factor completely x^{3} + 8y^{3}

Solution: Notice how the expression can be rewritten as (x) ^{3} + (2y) ^{3}, a sum of two
cubes. Recall that

**Example 4:** Factor completely x^{2} + 25

Solution: Recall that there is no special product for a sum of squares a^{2}
+ b^{2} .
In fact, this expression

cannot be factored; it is prime over the real numbers.

**IV. Factoring Binomials (x**^{2} + bx+ c )

^{2}+ bx+ c )

If possible, a binomial of this form must factor as (x + m)(x
+ n) , where m and n are
integers. Note that if this

is true:

We can see in the last statement that the numbers m and n that we seek must add up to b and multiply to c.

Example 5: Factor completely x^{2} + 7x + 12

Solution: We are looking for two numbers which will add up to 7 and multiply to
12. The numbers

(obtained by trial and error) are 3 and 4. Therefore,

**Example 6: **Factor completely x^{2} - 4x - 21

Solution: We are looking for two numbers which will add up to -3 and multiply to
-21. Note that since

the product is negative , the numbers must be different signs (one negative and
one positive). The numbers are

-7 and 3 (Note that if we had chosen 7 and -3, the numbers add up to +4. In
general, if the signs are

different , the larger number will have the same sign as b, the middle term).
Therefore,

**Example 7:** Factor completely x^{2} - 6x + 9

Solution: We are looking for two numbers which will add up to -6 and multiply to
9. Note that since

the product is positive, the numbers must be the same sign (both positive or
both negative). Since the middle

term is negative, the numbers must both be negative. The numbers are -3 and -3. Therefor

Note that we could have factored this using the special product ( a- b) ^{2}= a^{2 }- 2ab + b^{2}

**V. Factoring Trinomials (ax**^{2} + bx + c )

^{2}+ bx + c )

If possible, a trinomial of this form (where the leading coefficient is not 1)
must factor in the form

(mx + s)(nx + t) , where m, n, s, and t are all integers. It is possible to find these
numbers by trial and error;

however, such a method may at times be haphazard or tedious. The following is a
more straightforward

approach utilizing the technique of factoring by grouping.

First, we need to split the middle term bx into two terms which will allow us to
factor by grouping. To do this,

we use a similar technique as before, only now we look for two numbers whose sum
is b and whose product is

ac .

If you are curious as to why, see below:

As before, multiply the desired form out:

Now note that b = mt + ns and ac = (mn) (st) = (mt) (ns) . The numbers we are

looking for are mt and ns. How they split up into m,n, s, and t are determined
when we

factor by grouping.

**Example 8:** Factor completely 9x^{2 }- 3x
- 2

Solution: We first must find two numbers whose sum is -3 and whose product is (9)(-2)
= (-18). The

numbers are -6 and 3. Now we rewrite -3x as - 6x + 3x and factor by grouping:

(remember that you can check your answer by multiplying)

**Example 9**: Factor completely 6x^{2} - x - 15

Solution: We must first find two numbers whose sum is -1 and whose product is (6)
(-15) = -90. The

numbers are 9 and -10. Now we rewrite -x as 9x 1-0x and factor by grouping:

(Notice that both signs change in the last term)

If you are afraid of getting the middle terms in the wrong order , not to
worry...

the same as above.

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