Factoring Quadratics by inspection
Suggestion: Read the explanations and examples below and before
or after, as your like, for more examples, visit purplemath lessons on factoring
quadratics: the simple
case the hard case,
the weird
case. Multiple views are better than one. Mastery of the simple
case is required for mathematics 436. The remaining cases appear to be
optional.
Factoring by inspection is based on the backwards or indirect use
of the identity
[x+A][x+B] = x2+[A+B]x + AB.
that holds for all real numbers x, A and B. Why it holds is indicated as
follows.
Column Multiplication Method Explanation- A Digression
x + B
x +
A
×
x2+Bx
Ax + AB
x2+[A+B]x + AB.
Geometric Demonstration of [x+A][x+B] = x2+[A+B]x
+ AB
valid when x, A and B are all positive.
For x, A and B all positive, the area of the large rectangle is
[x+A][x+B] or the sum of the areas of the small rectangle. This implies
[x+A][x+B] = x2+[A+B]x + AB.
The condition that x, A and B all be positive can be removed if one uses
the distributive law twice to obtain this result
[x+A][x+B] = x[x+B] + A[x+B]
= [xx+xB] + [Ax+AB]
= [xx+Bx] + [Ax+AB]
= x2+ Bx +Ax + AB
= x2+[B+A]x + AB
= x2+[A+B]x + AB.
Factoring x2+bx + c by Inspection
a systematic approach.
Factoring by inspection of x2+bx + c where b and c
are integers uses the equation [x+A][x+B] = x2+[A+B]x + AB in
reverse. The methods works trying all pairs of integer
factors A and B for which AB = c to find a pair [A,B] with sum A+B
= b. This inspection method not guaranteed to work. Indeed it may fail.
That being said, if it works, it work quickly, and if fails, you need to
apply the other methods develop below to see if the quadratic can be
factored. A few examples follow.
First Factoring Example: x2+7x + 12 =
x2+[A+B]x + AB when A = 4 and B = 3 are two factors of 12. So
x2+7x + 12 = [x+A][x+B] = [x+4][x+3]
This application of x2+[A+B]x + AB.= = [x+A][x+B] to
factor a quadratic of the form x2+bx + c depended on our
finding integers A and B such that little b = A + B and little A B
= c = 12. Other possible values for [A,B] with A > B are
[12,1], [6,2], [4,3], [-3,-4], [-2,-6] and [-1,-12].
Remark: The identity
x2+7x + 12 = [x+4][x+3]
implies x2+7x + 12 = 0 when and only when x = -4 or x - 3. For
all other values of x, the product [x+4][x+3] will have non-zero
factors and so be nonzero.
Reflections Factoring by inspection in looking for integer roots of the equation
x2+Cx + D = 0
we want to express D as a product AB of two integers, with the
property that C =A+B is their sum. The latter equation C=A+B holds
for infinitely many pairs of integers [A,B]. But the equation D=AB only
holds for finitely many pairs, pairs obtained that may derived from
the prime factorization of the prime number D. The second and third factoring
by inspection examples show how. The approach here turns
factoring by inspection into a systematics process.
Second Factoring Example: Try to factor x2-3x - 18 by
inspection, that is by generation and inspection of all pairs of integers
[A,B] where A > B and AB = -18 for the property A+B = - 3.
Solution: We would like to find integers A and B such that AB =
-18 and A+B = -3.
The prime decomposition of -c= 36 = 3221
Now -36 = [-1]1 3221 =
AB with both A and B integers requires A must have the
form [-1]p3m2n where
3's-exponent m takes the values 0 or 1 or 2, and the 2's-exponent n
takes the values 0 or 1, and the [-1]'s exponent p is even or odd, say
value 0 or 1. Here B = -36/A will not have the same sign as A. It will
be negative when A is positive and positive when A is negative.
p
0,1
|
m
0,1,2
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n
0,1
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$A = [-1]^p 3^m 2^n $
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$ B = -\frac{18}A$
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$A+B$
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0
|
0
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0
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[-1]03020 = 1
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-18
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-17
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|
|
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1
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[-1]03021 = 2
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-9
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-7
|
|
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1
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0
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[-1]03120= 3
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-6
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-3
|
|
|
|
1
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[-1]03121 = 6
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-3
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3
|
|
|
2
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0
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[-1]03220 = 9
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-2
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7
|
|
|
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1
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[-1]03221 =18
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-1
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17
|
|
|
|
|
|
|
|
|
1
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0
|
0
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[-1]13020 = 1
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-18
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-17
|
|
|
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1
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[-1]13021 = 2
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-9
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-7
|
|
|
1
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0
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[-1]13120= 3
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-6
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-3
|
|
|
|
1
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[-1]13121 = 6
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-3
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3
|
|
|
2
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0
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[-1]13220 = 9
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-2
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7
|
|
|
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1
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[-1]13221 =18
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-1
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17
|
|
|
|
|
|
|
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The above table gives all possible combinations of integers A and B whose
product AB = -18. We may draw a tree diagram in place of filling a
table.
The A+B value of -3 appears in the third row where [A,B] =
[3, -6] So
x2-3x - 18 = [x+A][x+B] = [x+3][x-6]
Remark: The above method of finding A and B would not work for
quadratics of the form x2+bx -18 where b > 0 and little b
does not belong to the set of values for A + B above.
Shortcut: The A+B value of -3 appears in
also appears in the third row below the line where [A,B] = [-6, 3].
Since A and B have opposite signs and the roles of A and B are
interchangeable, and since we need only one solution of the equation
A+B = -3, we can adopt the convention that A > B in order to
eliminate the solution where the interchanging the values of A and B
gives a second solution. That convention A > B here
eliminates the need to generate the values
[-1]p3m2n where p is odd, or -1 since
the requirement A and B have opposite signs and A > B forces
A to be positive. If you do several examples, this shortcut will become
obvious: practice and numerical experience counts.
A further shortcut: When c is negative, the
inspection method for factoring x2+bx + c works by
generating all pairs of integers [A,B] for which A > 0 and AB = c
and seeing if |b| = A+B for one pair of the generated whole number
factors.
Third Factoring Example: Try to factor x2-9x + 20 by
inspection.
Solution: So if AB = 20 =
5122.[-1]2 with both A and B
integers number, A must have the form
[-1]p5m2n where the fives-exponent m
takes the values 0 or 1, and the twos-exponent n takes the values 0, 1 or
2, and the [-1] exponent p is even or odd, say value 0 or 1
|
p
|
M
|
N
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$A = [-1]^p5^M2^N $
|
$B = \frac{20}A$
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A+B
|
|
0
|
0
|
0
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[-1]05020 = 1
|
20
|
21
|
|
|
|
1
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[-1]05021 = 2
|
10
|
12
|
|
|
|
2
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[-1]05022 = 4
|
5
|
9
|
|
|
1
|
0
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[-1]051 20= 5
|
4
|
9
|
|
|
|
1
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[-1]05121 = 10
|
2
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12
|
|
|
|
2
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[-1]02251 = 20
|
1
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21
|
|
1
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0
|
0
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[-1]15020 = -1
|
-20
|
-21
|
|
|
|
1
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[-1]15021 = -2
|
-10
|
-12
|
|
|
|
2
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[-1]15022 = -4
|
-5
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-9
|
|
|
1
|
0
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[-1]151 20= -5
|
-4
|
-9
|
|
|
|
1
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[-1]15121 = -10
|
-2
|
-12
|
|
|
|
2
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[-1]12251 = -20
|
-1
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-21
|
Here the values of A increase in the third column but that does not
always occur. Your may draw a tree diagram in place of filling a
table. We see that the combination [A,B] = [4, 5] gives A+B = 9 and
AB = 20. So the foregoing implies
x2+9x + 20 = [x+A][x+B] = [x+4][x+5]
when [A,B] = [4, 5]. The same result follows from swapping the values of
A and B, that is taking [A,B] = [5,4] instead.
Remark: The above method of finding A and B would not work for
quadratics of the form x2+bx + 20 where b > 0 and little b
does not belong to the set of values for A + B above.
Shortcut: When c is positive, the inspection method for
factoring x2+bx + c works by generating all pairs of whole
numbers [A,B] for which AB = c and seeing if |b| = A+B for one pair of
the generated whole number factors.
Shortcut for third example: The value of C = 20 is positive. So if
A and B are integers with AB = C, we know that AB = C gives [-A][-B] = C
as well, and vice-versa. We also know that A and B will have the same
sign. The prime decomposition of 20 = 5122. So if
AB = 20 with both A and B being whole number, A must have the form
5m2n where the five exponent m takes the values 0
or 1, and the two exponent n takes the values 0, 1 or 2.
|
M
|
N
|
A = 5M2N.
|
B = 20/A
|
$A+B$
|
|
0
|
0
|
5020 = 1
|
20
|
21
|
|
|
1
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5021 = 2
|
10
|
12
|
|
|
2
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5022 = 4
|
5
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9
|
|
1
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0
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51 20= 5
|
4
|
9
|
|
|
1
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5121 = 10
|
2
|
12
|
|
|
2
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2251 = 20
|
1
|
21
|
We see that the coefficient of x in x2-9x + 20 is -9 is not in
the table of values for A+B, but its negative +9 is. So [A,B] = [-4, -5]
gives A+B = - 9 and AB = 20 as required.
Factoring ax2+bx + c by Inspection
a systematic approach and curiosity
Here want ax2+bx + c = [Ax+B][Cx+D]
Now want the product of the right side to equal the left hand side.
D +Cx
B+ Ax [times]
BD +BCx
ADx +
ACx2
BD +[BC +AD]x + ACx2 =a x2+bx + c
So we require BD = c and AC = a. So when a, b and c are integers, there
will be a product [Ax+B][Cx+D] = ax2+bx + c with A, B, C and D
integers provided there is a pair [A, C] of complementary factors of a;
another pair [B,D] of complementary factors of c such that BC +AD = b.
The foregoing suggests we generate and combine all complementary
factors of [A, C] of a with all complementary factors [B, D] of the
coefficient c, and see if the coefficient b appears as a value of BC +AD
for one of the combinations.
The equality C = a/A and D = c/B implies we need to evaluate the
expression
BC +AD = Ba/A + Ac/B
for all integer factors A of a and all integer factors B of c.
Lessons Elsewhere:
Explore this last thought further if you wish. I am going stop here
and recommend that you visit visit purplemath lessons on factoring
quadratics: the
simple case the hard case,
the weird
case. Multiple views are better than one. The foregoing may be the
hard or weird case - not part of mathematics 436. but the simple case
is.
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