Transcript Factor Quadratics

How to Factor Quadratics of the Form
2
ax
+ bx + c
1
The first rule of thumb in factoring is to factor out, if any, the
greatest common factor of all the terms, for example:
1. 2x2 + 6x – 8 = 2(x2 + 3x – 4)
2. 5x3 – 20x2 + 15x = 5x(x2 – 4x + 3)
On the other hand, what if the expression has no common
factors? For example, x2 + 4x + 2 and x2 + 4x + 3. Keep in
mind that not all quadratic expression can be factored. Our
first example x2 + 4x + 2 is not factorable (in this case, the
expression is called a prime). The second example, however,
x2 + 4x + 3 can be factored as (x + 1)(x + 3).
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There are a couple of ways we can determine whether a
quadratic expression is factorable or not. We will show you
some of those ways later. But first, we are going to show you
how to factor any quadratic expressions if they are indeed
factorable. We will divide these quadratic expressions into
two classes:
1. x2 + bx + c, that is, a = 1, and
2. ax2 + bx + c, where a  1
3
Let’s look at some of the examples in the first class and see
how to factor them:
1. x2 + 5x + 6
2. x2 – 7x + 10
3. x2 + 2x – 15
4. x2 – 6x – 16
If they are factorable, each of them must be factored as
(x )(x ) since that is the only way x2 can be factored.
How about the two numbers that should be also in the
parentheses—how do we get them? Those two numbers can
be obtained by asking, “What two numbers multiplied will
give me the last term?” Let’s repeat this again—what two
numbers multiply, not add. There are two pairs of numbers
that multiply to be 6, {1, 6} and {2, 3}, and for 10, {1, 10}
and {2, 5}, and for 15, {1, 15} and {3, 5}; whereas 16 has
three pairs: {1, 16}, {2, 8}, and {4, 4}. Which pair should we
use?
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1. x2 + 5x + 6
2. x2 – 7x + 10
3. x2 + 2x – 15
4. x2 – 6x – 16
Actually, it is quite easy to choose the right pair of numbers
that multiply to be the last term. This is how:
If the last term is positive, like in the first two examples, ask
what two numbers multiplied is the last term and their sum is
the coefficient of the middle term;
If the last term is negative, like in the last two examples, ask
what two numbers multiplied is the last term and their
difference is the coefficient of the middle term.
Here, you can ignore the sign of the coefficient of the middle
term. That is, just consider it is positive even if it is negative.
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1. x2 + 5x + 6
2. x2 – 7x + 10
3. x2 + 2x – 15
4. x2 – 6x – 16
If you understood what we said on the previous page, you
should have picked {2, 3}, {2, 5}, {3, 5}, and {2, 8} for our
four examples, respectively. That is, you should have written
something like the following:
1. x2 + 5x + 6 = (x 2)(x 3)
2. x2 – 7x + 10 = (x 2)(x 5)
3. x2 + 2x – 15 = (x 3)(x 5)
4. x2 – 6x – 16 = (x 2)(x 8)
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1.
x2 + 5x + 6 = (x 2)(x 3)
2.
x2 – 7x + 10 = (x 2)(x 5)
3.
x2 + 2x – 15 = (x 3)(x 5)
4.
x2 – 6x – 16 = (x 2)(x 8)
How should we determine the sign between the x and the number?
If the last term is positive, then both signs should be the same and they
follow the sign of the coefficient of the middle term, i.e., if that coefficient
is:
a) positive, then both signs should be “+,” the plus sign;
b) negative, then both signs should be “–,” the minus sign.
If the last term is negative and then one sign should be “+” and the other is
“–” and if the coefficient of the middle term is:
a) positive, then put “+” with the larger number and “–” with the
smaller number;
b) negative, then put “–” with the larger number and “+” with the
smaller number.
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If you followed what we said on the previous page, you should
have factored the quadratic expressions as follows:
1. x2 + 5x + 6 = (x + 2)(x + 3)
2. x2 – 7x + 10 = (x – 2)(x – 5)
3. x2 + 2x – 15 = (x – 3)(x + 5)
4. x2 – 6x – 16 = (x + 2)(x – 8)
If you think the rules on the previous two pages are more than you
can handle at this time, no problem—we have an easier way for
you to factor quadratic expressions like the ones above. All you
need to do is ask:
What are the two numbers that
• if I multiply, they will give me the last term, and
• if I add or subtract, they will give me the coefficient of the
middle term.
Again, ignore the signs of the the last term and the coefficient of
the middle term (we always can take care of the signs later).
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Let’s look at the following example:
x2 + 2x – 24
Recall the last term, 24, is obtained by multiplying. The only two
numbers that can add or subtract to be 2 in the middle term are 4
and 6. Therefore, we can forget about the other pairs that also
multiply to 24, namely, {1, 24}, {2, 12}, and {3, 8}. You will have
at least this:
x2 + 2x – 24 = (x
4)(x
6)
What about the signs? Since we obtain 24 by multiplying and its
sign is minus, so one sign must be “+” (plus) and the other must be
“–” (minus). So, just put the signs in.
x2 + 2x – 24 = (x – 4)(x + 6)
Are you puzzled? What happen if we put the “–” with the 6 and the
“+” with the 4—will that work too?
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The answer is no, it will not work. That’s why we should
always check whether we have the signs (if they are different)
in the right place or not. To see whether x2 + 2x – 24 is really
equal to (x – 4)(x + 6), some text books would suggest to
FOIL (x – 4)(x + 6) (i.e., to multiply it out) and see if it is
equal to x2 + 2x – 24.
F O I
L
(x – 4)(x + 6) = x2 + 6x – 4x – 24 = x2 + 2x – 24
Therefore, it’s the correct factoring, but a waste of time. We
obviously know the “F” will give us the first term, x2, and “L”
will give us the last term, –24. Therefore, we only need to
check “I” (the inner terms) and “O” (the outer terms).
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(x – 4)(x + 6)
–4x
+6x
Combine the inner and outer products by adding, the result is +2x,
which is exactly the middle term of x2 + 2x – 24. Therefore, the
factoring must be correct.
If, unfortunately, you have factored x2 + 2x – 24 as (x + 4)(x – 6),
you can still check it by the method mentioned above.
(x + 4)(x – 6)
+4x
–6x
Again, combine the inner and outer products, the result is –2x.
Although it’s not the +2x we want to have, it is exactly the
opposite. In this case, just switch the two signs, i.e., change
(x + 4)(x – 6) to (x – 4)(x + 6). It will be correct, with 100%
guarantee, you don’t have to check it again.
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Let’s look at the following example:
x2 + 13x + 30
In this case, we seem to have two pairs of numbers {2, 15} and {3, 10}
which will work, since each pair has a product of 30 and while the
difference of first pair is 13, the sum of the second pair is also 13. So
which pair? Why not write both out and see which is the correct pair?
Again, don’t write the signs yet.
(x 2)(x 15)
(x 3)(x 10)
We see 30 has a “+” sign. In order to have the “+” sign by multiplying,
either both signs have to be “+” or both be “–.” Since the 13 also has a “+”
sign, the signs must both be “+.” Put in the “+” signs, we can see only
(x + 3)(x +10) is correct:
(x + 2)(x + 15)
+2x
+15x
+17x
No!
(x + 3)(x + 10)
+3x
+10x
+13x
Yes!
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If you have been following what we have said so far, it is quite easy to
determine whether a quadratic expression is factorable or not. For
example, the following quadratic expressions are not factorable:
1. x2 + 6x – 24
2. x2 + 11x – 24
3. x2 + 5x + 24
24
1 24
2 12
3 8
4 6
Sum Difference
25
23
14
10
11
5
10
2
• The first one is not factorable since 6 (the coefficient of the middle
term) is neither the sum nor the difference of any desirable pairs of
factors of 24.
• The second one is also not factorable since the last term is –24; we
want the difference of the two factors of 24 to be 11, but 11 is not one
of such differences.
• The last one is also not factorable since the last term is +24; we want
the sum of the two factors of 24 to be 5, but 5 is not one of such sums.
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We have one more trick which we call the “closer factors”
trick. When the last term has more than one desirable pair of
factors, usually it is the pair of factors that are close to each
other or the pair(s) that are in the times table. For example, if
the last term is 42, it is likely the pair (which gives a product
of 42) is {6, 7}, more so than {3, 14}, {2, 21} and {1, 42}. If
the last term is 24, it is most likely a toss-up between {4, 6}
and {3, 8}, rather than a toss-up between {2, 12} and {1, 24}.
Although this trick will not be always the case, it is very
probable. It’s even more probable for quadratic expressions of
the form ax2 + bx + c where a  1.
We are going to discuss this type of quadratic expressions
next.
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Now, let’s look at the following example: 4x2 + 4x – 15.
4x2 + 4x – 15
In order to obtain the first term, 4x2 (by multiplying), we must use either 2x and 2x or x and
(2x 3)(2x 5)
4x. To get the last term, 15 (also by multiplying), (2x 1)(2x 15)
we must use either 3 and 5 or 1 and 15. This can
(x 3)(4x 5)
be quite a hassle, because it could be one of the
Which
(x 5)(4x 3) one?
six factor forms shown on the right. Which one?
We can’t really tell you which one is right, but
(x 1)(4x 15)
we can tell you that you should always try the
(x 15)(4x 1)
one with the factors that are closer to each other
first (recall the “closer factors” trick we have mentioned
(2x – 3)(2x + 5)
earlier). Therefore, try (2x 3)(2x 5) first. Check inner
and outer products again and see if it really works. The
–6x
inner product gives 6x while the outer gives 10x. Since
+10x
the15 of the last term has a “–” sign, so the signs must be
+4x
one “+” and the other “–”. Therefore, if we place the “+”
sign with the 5 and “–” with the 3, we will have +10x and –6x which will
combine to +4x. So, 4x2 + 4x – 15 = (2x – 3)(2x + 5).
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Let’s look at this example: 6x2 + 23x + 20.
If you are following the “closer factors” trick, you should have something
like (2x 5)(3x 4) or (2x 4)(3x 5). Let’s look at the second choice
(2x 4)(3x 5), notice that you can factor out a 2 from (2x 4). However,
the original expression 6x2 + 23x + 20
?
2 + 23x + 20 =
6x
(2x 4)(3x 5)
has no common factor. Therefore, it
Nothing can be factored
2 can be factored
can’t be (2x 4)(3x 5). We call this
out from here
out from here
the “no common factor” trick. That is,
Contradiction!
make sure that the numbers appearing
 6x2 + 23x + 20 can’t be factored as (2x 4)(3x 5)
in the same parenthesis have no
common factor if the given expression has no common factor.
Let’s go back to (2x 5)(3x 4) and see if it really works. We see that the
inner product gives 15x and the outer gives 8x and we
(2x + 5)(3x + 4)
see that the 20 of the last term has a “+” sign and the 23
+15x
of the middle term also has a “+” sign, so the signs must
be both “+”. So, 6x2 + 23x + 20 = (2x + 5)(3x + 4).
+8x
+23x
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Here is another example: 6x2 + 7x – 24.
Using the “closer factors” tactic, you would have (2x )(3x ) for the 6x2,
and the closest pair that gives a product of 24 is 4 and 6. In which parenthesis
should you put the 4 and the 6?
6x2 + 7x – 24 Nothing can be factored out from here
The answer is neither—don’t
be factored out from (2x 4) and
(2x 4)(3x 6) 23 can
put either number in. It is becan be factored out from (3x 6)
Can’t be this one!
cause it breaks the “no common
(2x 6)(3x 4) 2 can be factored out from (2x 6)
factor” rule. The next closest
Can’t be this one either!
pair that gives a product of 24 is (2x 8)(3x 3) 2 can be factored out from (2x 8) and
3 can be factored out from (3x 3)
3 and 8. Again, in which parenThis one is no good either!
thesis should you put the 3 and
(2x 3)(3x 8) Nothing can be factored out from here also
Hey, maybe it’s this one!
the 8? The answer is to put the 3
with 2x and the 8 with 3x, i.e., (2x 3)(3x 8).
The inner product of (2x 3)(3x 8) gives 9x while the outer
gives 16x. And it’s only possible to obtain the +7x (the middle term) if we assign the “+” sign with 8 and the “–” with 3
(see illustration on the right). Therefore, 6x2 + 7x – 24 =
(2x – 3)(3x + 8).
(2x – 3)(3x + 8)
–9x
+16x
+7x
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By using the “closer factors” and “no common factor” tricks wisely
(especially the latter one), it’s quite easy to determine whether the
quadratic expression is
“No Common Factor” Rule:
If an quadratic expression has no common factor in
factorable or not. For
the first place, no matter how you factor it, these
example, it’s not difficult
two numbers can’t have a common factor, so can’t these two.
to see that 6x2 + 11x + 48
is not factorable by using
(x )(x )
these two tactics.
48
If you have (3x )(2x ), notice that you can’t put in any pairs 6 8
of factors of 48 into (3x )(2x ), because one way or another 4 12
they all break the “no common factor” rule. Our last resort is 3 16
(6x )(x ). We can see that every pair break the rule except 2 24
1 48
the last pair 1 and 48. And we must put 1 with the 6x and 48
with x and see if it works. When we multiply the
(6x 1)(x 48)
inner terms and the outer terms, however, we will
1x
have 288x and 1x which can never become +11x of
288x
the middle term. So, 6x2 + 11x + 48 is not factorable.
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We can also determine whether a quadratic expression of the
form ax2 + bx + c is factorable or not by checking whether
b2 – 4ac (1) is a perfect square or not. If the result is a perfect
square (2), then the quadratic expression is factorable;
otherwise, it is not. Our example 6x2 + 11x + 48 has a = 6,
b = 11 and c = 48. In this case, however, b2 – 4ac = 112 –
4(6)(48) is negative—it can’t be a perfect square.
Therefore, 6x2 + 11x + 48 is not factorable.
Notes:
1. b2 – 4ac is called the discriminant of ax2 + bx + c.
2. A perfect square (including 0) is a number that is the square of some integer. Hence, the perfect squares are 0, 1, 4, 9,
16, 25, etc.
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Many areas in algebra requires the knowledge of factoring quadratics.
These include:
1. Solving most quadratic equations,
2. Solving rational equations,
3. Simplifying rational expressions, and last but not least,
4. Solving trigonometric, logarithmic, exponential and higher-degree
equations which are really quadratic equations in disguise.
Quadratic Equations
x2 + 5x – 6 = 0
2x2 + x – 3 = 0
3x2 + 4x = 7
4x2 – 3 = 4x
Rational Equations
x
1
3

 2
x  3 x  4 x  x  12
2
5
4


x 2  3x  4 x 2  1 x 2  5 x  4
Rational Expressions
Others
2 x 2  x  15 2 x 2  3x  14

2 x 2  9 x  10 x 2  2 x  3
x4 – 10x2 + 9 = 0
x 2  2 x  24 x 2  10 x  24

2 x 2  17 x  8 2 x 2  7 x  4
e2x + 7ex + 12 = 0
sin2x + 2sin x – 3 = 0
(log x)2 – 4log x – 5 = 0
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