Download The Mathematics 11 Competency Test

The Mathematics 11
Competency Test
Simple Trinomials as Products of
Binomials
Sometimes algebraic expressions of the form
ax2 + bx + c
(1)
where a, b, and c stand for actual numbers, can be written as a product of two binomials. This is
considered a worthwhile property primarily when the original coefficients a, b, and c, and any
numbers occurring in the resulting binomial factors are all whole numbers. When this situation
occurs, it gives us a way of factoring such trinomials, rewriting them as a product of two
somewhat simpler expressions.
Notice that
 x  a  x  b   x 2  a  b  x  ab
Thus, when we are given trinomial of the form
x2 + dx + e
where the coefficient of x2 is 1, we may be able to achieve this kind of factorization if we can find
two whole numbers, a and b, such that
a + b = d,
the coefficient of x in the original trinomial, and,
ab = e,
the constant term in the trinomial.
(You might think that the problem is now quite simple, since we have two unknowns, a and b, and
two equations, and so it’s just a matter of solving the system of equations. Unfortunately, there
are two problems with this: most importantly, we are only interested in solutions which are whole
number values, and also, the second equation here is not a linear equation. This doesn’t mean
you can’t solve the system of equations given above in specific instances. However, it usually
means part of that process involves solving a so-called quadratic equation, and that is a topic not
covered in full detail in the BCIT Mathematics 11 Competency Test. In the examples below, we
will demonstrate a systematic inspection method which produces the values a and b required
here when they actually exist.)
Example 1: Factor x2 + 5x + 6 if possible.
solution:
A minute’s examination indicates that these three terms contain no common monomial factors.
However, since this expression has the pattern of expression (1) above with the coefficient of x 2
equal to 1, there is the possibility of factoring this expression into the product of two binomials,
(x + a)(x + b). For this to be possible, we need to find two numbers, a and b, such that
a+b=5
and
ab = 6.
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 1 of 9
We could just sit for a while and wait for inspiration, or we could test out a few guesses.
However, we can be a bit more systematic. The more restrictive condition is that ab = 6. There
aren’t many possibilities. We’ll make a little table listing all of the pairs of whole numbers whose
product is 6:
a
1
2
-1
-2
b
6
3
-6
-3
a+b
7
5
-7
-5
(You might think that a = 6, b = 1, for instance, should also be in our table. However, swapping
values of a and b just amounts to reversing the order of the factors in the final product, and so
does not give any new possibilities. It is important to consider both positive and negative factors
which multiply to give +6 in this case, however.)
From this table, we see that there are only four pairs of whole numbers which multiply to give +6.
We’ve also listed the sum of the two numbers in each case, and you can see that one of these
pairs of values does sum to +5. Thus, a = 2 and b = 3 seems to satisfy the requirements here.
Checking,
(x + 2) (x + 3) = (x + 2)(x) + (x + 2)(3)
= x(x + 2) + 3(x + 2)
= x2 + 2x + 3x + 6
= x2 + 5x + 6
which is identical to the original expression in this example. Hence, in factored form we can write
x2 + 5x + 6 = (x + 2) (x + 3)
(Note that if the table above containing pairs of whole numbers with a product of +6 had
contained no row in which the two numbers summed to +5, this would be proof that the type of
factorization we were attempting could not be achieved. Thus, this systematic inspection method
provides a factorization of a trinomial like this when such a factorization exists, and demonstrates
that no such factorization is possible when that is the case.)
Example 2: Factor x2 – 2x – 35 as completely as possible.
solution:
The terms of this expression contain no common monomial factors. However, since this
expression has the form of a trinomial in x, with the coefficient of x2 being 1, factoring into the
product of two binomials, (x + a)(x + b) may be possible. We would need to find values for a and
b satisfying
a + b = -2
and
ab = -35
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 2 of 9
As in the previous example, we begin by making a table of all pairs of whole numbers which
multiply to give, in this case, -35. The easiest way to make sure our list is complete is to make
sure that the value of ‘a’ covers all positive and negative whole numbers which divide evenly into
-35. The value of ‘b’ in each case is obvious, because we need to have ab = -35.
a
1
5
7
35
-1
-5
-7
-35
b
-35
-7
-5
-1
35
7
5
1
a+b
-34
-2
2
34
34
2
-2
-34
It appears that there are eight candidate pairs of whole numbers that multiply to give -35.
(However, if you examine the table carefully, you’ll see that only four of these pairs of values are
unique. The last four rows essentially duplicate the first four rows, amounting to swapping the
order of the two possible factors in the eventual product. So, we’ve done twice as much work
here as was really necessary.)
To achieve the proposed factorization, one of these pairs of numbers must sum to -2. From this
table, this is seen to occur when a = 5 and b = -7. Thus, it appears that
x2 – 2x – 35 = (x + 5)(x – 7)
is the required answer.
Checking,
(x + 5)(x – 7) = x(x – 7) + 5(x – 7)
= x2 -7x + 5x – 35
= x2 - 2x – 35
confirming our answer.
Example 3: Factor as completely as possible: x2 + 8x + 10.
solution:
There are no common monomial factors here, but the expression has the form of a trinomial of
the type considered in the previous two examples. Thus, a possible factorization has the form of
a product of two simple binomials, (x + a)(x + b), if we can find two numbers, a and b, satisfying
a+b=8
and
ab = 10
Setting up the usual table of pairs of whole numbers with a product of +10, we get
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 3 of 9
a
1
2
-1
-2
b
10
5
-10
-5
a+b
11
7
-11
-7
(Here, we listed values of ‘a’ which divide evenly into +10, only for |a| < |b|, realizing from the
previous example that to also include rows in the table for a = 5 and a = 10 will just duplicate
the information in these first four rows.)
To get this factorization to work, we need to find a pair of these whole numbers which sum to 8.
However, the table contains all possibilities, and none of the rows give a sum of 8. So, we must
conclude that the given trinomial cannot be factored.
(For more on this expression, see Example 15 in the document on the multiplication and division
of square roots, later in these notes.)
In the special situation that a = b, we get the form
 x  a  x  a    x  a 
2
 x 2  2ax  a 2
Thus, a trinomial in which the coefficient of x2 is 1 is the square of a binomial if
(i) the constant term is a perfect square (of a whole number)
and
(ii) the coefficient of the x term is double the square root of the constant term.
You don’t really need to memorize this formula as a special case, because the more general
method described above will also work in this case. You’ll just find that the two whole numbers, a
and b, that you get from the analysis will be equal.
Example 4a: Factor x2 + 10x + 25 as much as possible.
solution:
The constant term here is a perfect square: 25 = 52. Its square root is 5. But 2 x 5 = 10, which
happens to be the coefficient of x. So, we can write
x2 + 10x + 25 = (x + 5)2
You can check that this is correct by multiplying the right-hand side to remove the brackets, to
confirm that the result obtained is the expression on the left-hand side.
Example 4b: Factor x2 - 10x + 25 as much as possible.
solution:
This expression is very similar to the one in Example 4a. The constant term is a perfect square,
with square root of 5. In this case, the coefficient of the x term is -10 = 2 x (-5), so the trinomial
has the form in the box above, but with a = -5. Thus, we conclude that
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 4 of 9
x2 - 10x + 25 = (x – 5)2
This is easily verified:
(x – 5)2 = (x – 5)(x – 5)
= x(x – 5) + (-5)(x – 5)
= x2 - 5x – 5x + (-5)2
= x2 – 10x + 25
as required.
Example 4c: Factor x2 - 10x - 25 as much as possible.
solution:
At first glance, this trinomial seems to have the same form as the trinomials in both Examples 4a
and 4b. However, matching this trinomial with the pattern shown in the box just above is not
correct. Notice that in the template formula in the box, all terms are connected by ‘+’ signs.
Thus, to match the trinomial in this example with that template, we need to write
x2 - 10x – 25 = x2 - 10x + (-25)
Now we see that the constant term must be considered to be -25, not 25, and so it is not a perfect
square. Thus, this trinomial cannot possibly match the pattern to potentially be equivalent to the
square of a binomial.
You could still try to factor this trinomial using the more general method, to see if it can be
factored into a product of the form (x + a)(x + b). The familiar table of possible values of a and b
is:
a
1
5
-1
-5
b
-25
-5
25
5
a+b
-24
0
24
0
Since none of the pairs of values which multiply to give -25 also sum to give -10, we conclude
that no factorization of any form of this trinomial is possible.
Example 4d: Factor x2 - 10x + 25 as much as possible. Use the general method for factoring
trinomials.
solution:
This is the same trinomial as considered in Example 4b above. In this case, we will not make use
of the special form of the coefficients to recognize that it is equivalent to the square of a binomial.
Instead, we’ll confirm that the general method for factoring simple trinomials will automatically
generate that result.
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 5 of 9
So, if this trinomial is factorable, it will be into the form (x + a)(x + b), where
a + b = -10
and
ab = 25
The possible pairs of whole numbers which multiply to 25 are listed in the table
a
1
5
-1
-5
b
25
5
-25
-5
a+b
26
10
-26
-10
The fourth row of this table gives a sum of -10. Thus using a = -5 and b = -5 (that is, a = b = -5),
we get
x2 - 10x + 25 = (x + (-5)) (x + (-5)) = (x – 5)(x – 5) = (x – 5)2
as before.
The Most General Trinomial
Both of the above methods work only for trinomials in which the coefficient of the x 2 term is 1.
More general trinomials can sometimes also be factored. Note that
ax  b cd  d   ac x 2  (ad  bc )x  bd
So, a general trinomial (one in which the coefficient of x 2 is a whole number different from 1) can
be factored into the form of a product of two binomials, (ax + b)(cx + d), if we can find four whole
numbers, a, b, c, and d, satisfying
ac = the coefficient of x2
ad + bc = the coefficient of x
and
bd = the constant term
Analyzing these conditions can require quite a lengthy process, depending on the specific values
of the various coefficients, because of the large number of possibilities that must be examined,
and the complexity of the second condition.
The easiest approach is to make a table of sets of whole number values of a, b, c, and d, which
satisfy the first and third conditions above. These are the most restrictive conditions, and so will
narrow down the possibilities to be examined as much as possible. For each quadruple of values
in the table, the value of ad + bc would then be calculated to determine which quadruple, if any of
the possibilities, satisfies the second condition. If such a set of values is found, the factorization
follows according to the pattern shown in the box above. If no such set of values is found, the
trinomial cannot be factored in this way.
We illustrate this method with a simple example.
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 6 of 9
Example 5: Factor 6x2 + 13x – 5 as much as possible.
solution: The three terms here contain no common monomial factors, so the only possibility is to
try to factor this expression into the product of two binomials. Since the coefficient of x 2 is not
equal to 1, we must seek a factorization of the form (ax + b)(cx + d). The numbers in this product
must satisfy
ac = 6
(coefficient of x2)
bd = -5
(constant term)
ad + bc = 13
(coefficient of x).
Since ac = 6 is a positive value, we know that a and c must have the same signs. The positivevalued possibilities for a and c are
a=1
a=6
a=2
a=3
and c = 6
and c = 1
and c = 3
and c = 2
We don’t need to consider negative values of a and c as long as we include both positive and
negative possibilities for b and d. The list of four possible pairs of values for a and c above might
appear at first to duplicate each actual possibility. However, since a occurs in the factor with b,
and c occurs in the factor with d, the situation a = 1 and c = 6 is actually different than the
situation with a = 6 and c = 1.
The condition bd = -5 restricts possible values of b and d to the combinations
b=1
b = -1
b=5
and
and
and
d = -5
d=5
d = -1
b = -5
and
d = 1.
and
Note that our list includes the positives and negatives of all values of b which divide evenly into
-5. This is necessary to ensure that our analysis includes all possible combinations of whole
numbers that might work here.
The four possible pairs of values for a and c can now be matched with each of the four possible
pairs of values for b and d, giving a total of 16 possibilities to be check. (You can see that when
the coefficient of x2 and the constant term of the trinomial have a lot of whole number factors, the
combined number of possibilities to be examined at this stage can become very large, so that this
factorization can become a very tedious job – and often the end result is the conclusion that the
trinomial cannot be factored!)
Anyway, setting the analysis up in a table gives:
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 7 of 9
a = 1, c = 6
a = 6, c = 1
a = 2, c = 3
a = 3, c = 2
a
1
1
1
1
6
6
6
6
2
2
2
2
3
3
3
3
b
1
-1
5
-5
1
-1
5
-5
1
-1
5
-5
1
-1
5
-5
c
6
6
6
6
1
1
1
1
3
3
3
3
2
2
2
2
d
-5
5
-1
1
-5
5
-1
1
-5
5
-1
1
-5
5
-1
1
ad + bc
1
-1
29
-29
-29
29
-1
1
-7
7
13
-13
-13
13
7
-7
 we can stop here, since we’ve
found a set of values that work!
This row has a, b, c, and d
satisfying ab = 6, bd = -5 and
ad + bc = 13, as required.
Thus, it appears that
6x2 + 13x – 5 = (2x + 5)(3x – 1)
Checking,
(2x + 5)(3x – 1) = (2x + 5)(3x) + (2x + 5)(-1)
= (3x)(2x + 5) +(-1)(2x + 5)
= (3x)(2x) + (3x)(5) + (-1)(2x) + (-1)(5)
= 6x2 + 15x – 2x - 5
= 6x2 + 13x – 5
as required. Our proposed factorization of the original trinomial has been verified.
Two, no three, final comments with regard to the method demonstrated with the example above:
(i) It is logically ok to sort of randomly test various possibilities in hope of finding an acceptable
combination of values that might work without having to test all of the possibilities in the table. It
is important to set the table up systematically however. If you just try random combinations of
values of a, b, c, and d, without some way of keeping track of which combinations you’ve tested,
you’ll end up going in circles, almost certainly. Of course, the only way to conclude that a
trinomial of this type cannot be factored is to make sure you test all possible combinations of
values of these four constants. (Yes, if you know how to solve a quadratic equation, you can
achieve this factorization or demonstrate that no factorization is possible with quite a lot less
work.)
(ii) You can see from the table of test cases above that we still have duplication. We leave it up
to you to figure out how to avoid at least some of the duplication that occurred above, if you are
interested. (Hint: the product (ax + b)(cx + d) is identical to the product (cx + d)(ax + b).)
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 8 of 9
(iii) This is really a very tedious type of problem to solve even under the best of circumstances.
There are more powerful and efficient methods for finding the factors of a trinomial which get
around the need to enumerate large sets of possible factorizations, but these methods are also
more complicated and beyond the scope of the BCIT Mathematics 11 Competency Test.
David W. Sabo (2003)
Simple Trinomials as Products of Binomials
Page 9 of 9