January 26th
January 26th
Monomial Factors
The first stage of factoring an algebraic expression involves
the identification of any factors which are monomials. We will
describe the process by working through several examples in
detail. Sometimes the expression to be factored is simple enough
to be able to use straightforward inspection.
Example 1:
Factor the expression: 2xy + 6yz.
solution:
We see that
2xy = 2 · x · y
and
6yz = 2 · 3 · y · z
We can quickly recognize that both terms contain the factors 2
and y in common. Thus
2xy + 6yz = 2y(x + 3z).
The right hand side here is in factored form because it is a
single term only. As it is written, it does not consist of two or
more parts which are connected by plus or minus signs. It is just
a single term which is the product of three factors in this case.
That the factored expression above is mathematically
equivalent to the original expression is easily verified by
multiplication.
2y(x + 3z) = (2y)(x) + (2y)(3z)
= 2yx + 6yz
= 2xy + 6yz.
In fact, you can always verify that a factorization is correct
by remultiplication to confirm that the original expression is
regenerated.
Example 2:
Remove all common monomial factors from: 16x^{ 3}
– 24x^{ 2} + 84x.
solution:
This expression is a bit more complicated than the one in the
previous example. With it, we can begin to implement the
systematic method that should be used in all but the very
simplest cases where the factorization is immediately obvious.
(In fact, you should probably always do the systematic method,
because that is really the only way to make sure that what you
thought was an easytosee simple factorization really is
complete and correct!)
We start by selecting any term in the expression  it
doesn’t matter which one we pick, so we will pick the first
one here: 16x^{3}.
Enumerate the factors which make up this term:
16x ^{3} = 2^{ 4} · x^{ 3}
This means that any monomial factor common to all three terms
in the original expression must either be a power of 2 or a power
of x. This is because monomial factors common to all three terms
must be factors of each individual term, hence they must be a
factor of the first term (or whichever one you picked). So, all
we have to do is check what power of 2 is common to all three
terms, and what power of x is common to all three terms.
It’s probably easiest at this stage to write out the
factorizations of each term for easy reference:
16x ^{3} = 2^{ 4} · x^{ 3}
24x^{ 2} = 2 ^{3 }· 3 · x^{ 2}
and
84x = 2^{ 2} · 3 · 7 · x^{ 1}
(The method for resolving whole numbers into products of prime
factors has been described in an earlier document in these
notes.) From this we see that
 all three terms contain a factor of 2^{ 2}, so
this is a factor of the entire expression
 all three terms contain a factor of x^{ 1} or x,
so this is a factor of the entire expression
 the above two bullets in this list cover all possible
factors of the first term, and so cover all possible
monomial factors of the entire expression.
Thus, 2^{ 2} x = 4x is the most extensive monomial
factor common to all three terms of the original expression. We
can write
16x^{ 3} – 24x^{ 2} + 84x = 4x (4x^{ 2}
– 6x + 21)
The trinomial in brackets is obtained by removing the factor
4x from each of the three terms in the original expression:
16x^{ 3} ÷ 4x = 4x^{ 2}
24x^{ 2} ÷ 4x = 6x
84x ÷ 4x = 21
The expression on the right above is the required factored
expression here. The terms of the trinomial in the brackets
contain no further common monomial factors, and so all of the
common monomial factors have been “factoredout” as
required. You should take a minute here to verify that if you
multiply to remove the brackets from the factored expression, you
get precisely the original expression back again.
