Video Transcript
In this video, we will learn how to
factor trinomials into a product of two binomials.
We begin by recalling the
definitions of monomials, binomials, and trinomials. A monomial is a product of numbers
and powers of variables. A binomial expression is the sum or
difference of two monomials. A trinomial expression is the sum
or difference of three monomials. An example of a monomial expression
is negative five 𝑥 squared 𝑦. An example of a binomial expression
is three 𝑥 squared plus seven. An example of a trinomial
expression is two 𝑎 squared minus two 𝑎𝑏 plus three 𝑏.
When we list the factors of a
number, we can write the number as a product of its factors. For example, we may write 20 equal
to two multiplied by 10. The same principle is true when we
factor algebraic expressions. The focus of this video is writing
trinomials as the product of two binomial factors.
In general, when we multiply two
binomials, we initially obtain four terms, created by multiplying each term in one
binomial by each term in the other. If the two binomials have the same
algebraic structure, then we are able to combine one pair of like terms, leading to
a trinomial. Each of the examples we consider in
this video are of this type. We first demonstrate the process of
factoring an expression with a common binomial term that is already in a partially
factored form.
Fully factor the expression 𝑥
multiplied by 𝑥 plus three plus two multiplied by 𝑥 plus three.
Upon inspection, we observe
that the two parts of this expression share a common binomial factor of 𝑥 plus
three. We can, therefore, factor by
this shared binomial. In the first part of the
expression, this binomial is multiplied by 𝑥. And in the second part, it is
multiplied by two. Hence, overall, it is
multiplied by 𝑥 plus two. And so, 𝑥 multiplied by 𝑥
plus three plus two multiplied by 𝑥 plus three is equal to 𝑥 plus two
multiplied by 𝑥 plus three. This expression cannot be
factored any further as the two terms in each binomial do not share any common
factors other than one.
We now consider how to factor a
quadratic expression of the form 𝑥 squared plus 𝑏𝑥 plus 𝑐 into the product of
two binomials. Factoring is the reverse process of
distributing parentheses or expanding brackets. Consider the expansion of the
product of the binomials 𝑥 plus five and 𝑥 plus three. We begin by distributing the first
set of parentheses over the second, giving us 𝑥 multiplied by 𝑥 plus five plus
three multiplied by 𝑥 plus five. Distributing each set of
parentheses, we have 𝑥 squared plus five 𝑥 plus three 𝑥 plus 15. And collecting like terms, this
simplifies to 𝑥 squared plus eight 𝑥 plus 15.
We can observe that the constant
term in the trinomial expression is the product of a constant terms in the two
binomials. 15 is equal to three multiplied by
five. The coefficient of 𝑥 in the
trinomial is the sum of the constant terms in the two binomials. Eight is equal to three plus
five. And in the penultimate line of our
solution, the 𝑥-term is written as the sum of two terms with these coefficients,
five 𝑥 plus three 𝑥. This suggests a process we can
follow to work in the opposite direction and factor an expanded quadratic of the
form 𝑥 squared plus 𝑏𝑥 plus 𝑐 into the product of two binomials. It is important to note that not
all quadratics of this form can be factored, so the process that follows is only
applicable to those that are factorable.
Firstly, we list the factor pairs
of the constant 𝑐. If 𝑐 is positive, the two numbers
will have the same sign, whereas if 𝑐 is negative, the two numbers will have
opposite signs. Look for a factor pair that with
the correct combinations of signs sum to the coefficient of 𝑥, i.e., 𝑏. Rewrite the middle term in the
trinomial as the sum of two terms with coefficients equal to the factor pair
found. Separate the new four-term
expression into two binomials and factor each. Look for a shared binomial factor
to factor the entire expression by. We will demonstrate this process of
factoring a quadratic of the form 𝑥 squared plus 𝑏𝑥 plus 𝑐 in our next
example.
Factor 𝑥 squared minus eight
𝑥 minus 20.
To factor this quadratic
expression, we wish to write it as the product of two binomials. We will do this by first
rewriting the middle term as the sum of two terms with coefficients whose sum is
the coefficient of 𝑥 and whose product is the constant term. We first consider the factor
pairs of 20. As the product of the two
numbers must be negative 20, the two numbers must have different signs. If we choose the second factor
pair of two and 10 and choose the two to be positive and the 10 to be negative,
then the sum of these two numbers is two plus negative 10, which is equal to
negative eight as required.
We then rewrite the trinomial
with the middle term expressed as the sum of two terms with coefficients of two
and negative 10, that is, 𝑥 squared plus two 𝑥 minus 10 𝑥 minus 20. Separating this four-term
expression into two binomials and factoring gives us 𝑥 multiplied by 𝑥 plus
two minus 10 multiplied by 𝑥 plus two. Finally, we factor the entire
expression by the shared binomial factor of 𝑥 plus two to give 𝑥 plus two
multiplied by 𝑥 minus 10. This is the fully factored form
of 𝑥 squared minus eight 𝑥 minus 20.
We have seen in this example how to
factor quadratics of the form 𝑥 squared plus 𝑏𝑥 plus 𝑐 into the product of two
binomials. Such quadratics, in which the
coefficient of 𝑥 squared is equal to one, are known as monic quadratics. We will now consider the more
general case of how to factor a nonmonic quadratic of the form 𝑎𝑥 squared plus
𝑏𝑥 plus 𝑐, where 𝑎 is not equal to zero, one, or negative one. We will demonstrate this process
through an example.
Consider the product of the
binomials three 𝑥 minus two and two 𝑥 plus five.
Using the distributive property, we
have three 𝑥 multiplied by two 𝑥 plus five minus two multiplied by two 𝑥 plus
five, which simplifies to six 𝑥 squared plus 15𝑥 minus four 𝑥 minus 10 and in
turn to six 𝑥 squared plus 11𝑥 minus 10. Now we consider the reverse of
this, expressing the trinomial six 𝑥 squared plus 11𝑥 minus 10 as the product of
two binomials. We observe that in the first line
of our working, the two parts of the expression had a shared binomial factor of two
𝑥 plus five.
To work in the other direction, we
first need to rewrite a trinomial as an expression involving four terms so that we
can then separate the resulting expression into one that contains two binomials and
factor each separately. We begin by looking for two numbers
whose sum is the coefficient of 𝑥, in this example 11, and whose product is equal
to the product of the coefficient of 𝑥 squared and the constant term, in this case
six multiplied by negative 10, which is equal to negative 60. These two numbers are 15 and
negative four. We then rewrite the trinomial as a
four-term expression, separating the 𝑥-term into the sum of two terms with these
coefficients.
Next, we split this expression into
two binomials and factor each binomial separately, giving us three 𝑥 multiplied by
two 𝑥 plus five minus two multiplied by two 𝑥 plus five. This reveals a common binomial
factor of two 𝑥 plus five, which can subsequently be factored to give our solution
two 𝑥 plus five multiplied by three 𝑥 minus two. This method of factoring is the
reverse process of expanding the brackets that we see on the right-hand side of the
screen. The method for factoring a monic
quadratic that we met earlier is, in fact, a special case of this method, in which
the product of 𝑎 and 𝑐 is equal to 𝑐 because 𝑎 equals one.
The examples we have considered so
far have concerned trinomials involving only a single variable. In our final example, we will
consider how to factor a trinomial involving two variables.
Factorize fully 48𝑚 to the
fourth power plus 48𝑚 squared 𝑛 minus 15𝑛 squared.
We begin by observing that the
coefficients of all three terms are multiples of three. Hence, we can factor the entire
trinomial by three, which gives us three multiplied by 16𝑚 to the fourth power
plus 16𝑚 squared 𝑛 minus five 𝑛 squared. Now consider the structure of
the trinomial. The first term involves 𝑚 to
the fourth power, which is equal to 𝑚 squared all squared. And the third term involves 𝑛
squared. The central term involves a
product of 𝑚 squared and 𝑛. This suggests that the factored
form of the trinomial is 𝐴𝑚 squared plus 𝐵𝑛 multiplied by 𝐶𝑚 squared plus
𝐷𝑛, for values of 𝐴, 𝐵, 𝐶, and 𝐷 to be determined.
We now need to find two numbers
whose sum is equal to the coefficient of 𝑚 squared 𝑛, in this case 16, and
whose product is equal to the product of the coefficients of the first and last
terms. 16 multiplied by negative five
is equal to negative 80. The factor pairs of 80 are as
shown. As the product should be
negative 80, we require a factor pair with opposite signs such that their sum is
equal to 16. The correct pair are 20 and
negative four. Rewriting the second term in
the trinomial as the sum of two terms with these coefficients gives 16𝑚 to the
fourth power plus 20𝑚 squared 𝑛 minus four 𝑚 squared 𝑛 minus five 𝑛
squared.
Separating this four-term
expression into two binomials and factoring each separately gives four 𝑚
squared multiplied by four 𝑚 squared plus five 𝑛 minus 𝑛 multiplied by four
𝑚 squared plus five 𝑛. Hence, the fully factored form
of the trinomial is three multiplied by four 𝑚 squared minus 𝑛 multiplied by
four 𝑚 squared plus five 𝑛.
Let us finish by recapping the key
points from this video.
To factor a nonmonic quadratic of
the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where 𝑎 is not equal to zero, one, or
negative one, we perform the following steps. List the factor pairs of 𝑎𝑐. Look for a factor pair that with
the correct combination of signs sum to the coefficient of 𝑥, noting that if 𝑎𝑐
is positive, the two numbers will have the same sign, whereas if 𝑎𝑐 is negative,
the two numbers will have opposite signs. Rewrite the middle term in the
trinomial as the sum of terms with coefficients equal to the factor pair found. Separate the new four-term
expression into two binomials and factor each. Look for a shared binomial factor
to factor the entire expression by.
Factoring monic quadratics of the
form 𝑥 squared plus 𝑏𝑥 plus 𝑐 is a special case of the above, where 𝑎 is equal
to one. And hence, 𝑎𝑐 is equal to 𝑐. To factor a two-variable trinomial,
first consider its structure and identify the structure of the binomial factors. The coefficients can be found using
the same method used for nonmonic quadratics. Note that some trinomials can be
factored into the product of more than two terms by first taking out a common
factor.
Whilst we will not cover it in this
video, the techniques we have encountered here can be applied to problems in other
areas of mathematics, such as geometry or real-world problems.
