Factoring Quadratics by inspection

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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

x + B
x + A               (times)
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.

Second 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)^p3^m2ⁿ 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.

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)^p3^m2ⁿ 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 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)^p5^m2ⁿ 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

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  5^m2ⁿ where the five exponent m takes the values 0 or 1, and the two exponent n takes the values 0, 1 or 2.

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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