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Factor by Grouping Short-cut. ALWAYS CHECK FOR A GCF OF THE POLYNOMIAL 1ST!
ax 2 + bx + c The key is to find the factors that satisfy the following rule:
_____ * ______ = ac
_____ + ______ = b
These two factors are the O and I from FOILing. Next factor by grouping.
Example: 10 x + 7 x − 12
2
_____ * ______ = ac = 10 * (–12)
_____ + ______ = b = 7
ODD RULE
Since b = 7 is odd, there is only one way for the sum of two numbers to be odd.
Even + odd. Find all the possible odd factors in 10(-12). 10(12) = 2*5*4*3.
Regroup the factors with the odd factors by themselves.
3*______, 5*_______, and 15*_______. Remember odd * odd = odd. 3 * 5 = 15
3 * 40,
5 * 24
,
15 * 8. These are the only possible odd and even factors that will
work besides 1 and ac. 1 and ac is only used if b is very large. 15 and 8 are the two factors that
can subtract to equal 7. Since we subtracted, make the 8 a negative 8.
_15_ * _–8_ = ac = 10 * (–12)
_15_ + _–8_ = b = 7
Replace 7x with 15x – 8x and factor by grouping. Doesn’t matter if it is -8x +15x.
10 x 2 + 15 x − 8 x − 12
10 x 2 − 8 x + 15 x − 12
2 x(5 x − 4 ) + 3(5 x − 4 )
5 x(2 x + 3) − 4(2 x + 3)
(2 x + 3)(5 x − 4)
(5 x − 4)(2 x + 3)
Notice that the first binomial created in step two of each set of work are the binomials of the
answer. Concentrate on only the left side of each factor by grouping and you notice that the
leading coefficient 10 is used twice with 15 and -8. We will always factor out the factor of x on
the left side, so let’s leave that x out of the left side. This would give use the two binomials
(10 x + 15) and (10 x − 8) . If you remove the GCF of each binomial, you are left with the answer.
(2 x + 3)(5 x − 4) The work can be shortened to the following:
10 x 2 + 7 x − 12
_15_ * _–8_ = ac = 10 * (–12)
(10 x + 15) (10 x − 8)
(10 x + 15) (10 x − 8)
5
2
(2 x + 3)(5 x − 4)
_15_ + _–8_ = b = 7
The 5 and 2 that you are dividing out are the GCF’s from
the work above and the extra 10 we used is now gone.
Example: 8 x 2 − 14 x − 15
_____ * ______ = ac = 8 * (–15)
_____ + ______ = b = –14
Since b = -14 is even, there are only two ways for the sum of two numbers to be even.
Even + even or odd + odd. Since the 8 and 15 are not both odd, you must find two even factors.
If ac are both odd, then look for two odd factors.
Find all the possible even factors in 8(-15). 8(15) = 2*4*5*3.
Regroup the factors with the even factors separated. ( 2 * ____ ) and ( 4 * ____ ).
With the even factor in each number you will get even numbers.
Swap the odd factors into the blanks until you get the numbers that work.
( 2 * _5_ ) and ( 4 * _3_ ). 10 and 12
( 2 * _3_ ) and ( 4 * _5_ ). 6 and 20 This works…6 – 20 = –14.
8 x 2 − 14 x − 15
_6_ * _–20_ = ac = 8 * (–15)
_6_ + _–20_ = b = –14
(8 x + 6) (8 x − 20)
(8 x + 6) (8 x − 20)
2
4
Notice that 2 * 4 is 8 and that is the extra 8 that we put into the
binomials! Removing them as the GCF removes the extra 8.
(4 x + 3)(2 x − 5)
Another approach. Since we know that both numbers have to be even, the numbers must have
a factor of 2 in them. This way we factor out the 2 from the second line, including b = - 14.
_____ * ______ = ac = 8 * (–15) = 2 * 2 * 2 * 3 * 5
2 (___)_ + 2 (___)_ = b = –14 = 2(– 7 )
_6_ * _–20__ = ac = 8 * (–15)
2(3) + 2(–10) = b = –14
Now the factors 2, 3, & 5
must create two numbers that
+ or – to be – 7! Odd rule.
2*5 is 10…-10 + 3 = -7
Example: 3 x 2 − 34 x + 63
_____ * ______ = ac = 3 * 63
_____ + ______ = b = – 34
Since b = – 34 is even, there are only two ways for the sum of two numbers to be even.
Even + even or odd + odd. Since the 3 and 63 are both odd, you must find two odd factors.
3 * 63 = 3 * 9 * 7 = 3 * 3 * 3 * 7. List all possible odd factors. Remember odd * odd = odd.
3 * ____, 7 * _____, 9 * _____, 21 * _____, 27 * ____, and 63 * ____. It appears that we may
have many possible groups, but watch what happens to the second numbers in our grouping.
3* _63_, 7* _27_, 9 * _21_, 21 * _9_, 27 * _7_, and 63 * _3_. The second numbers are the first
factors in reverse order.
Recommend to fill the blanks as you start your listing.
When the second number begins to repeat, stop! The choices are 3 & 63, 7 & 27, and 9 & 21.
Which pair adds or subtracts to be – 34? 7 + 27 = 34. Change them to negatives – 7 – 27 = – 34.
3 x 2 − 34 x + 63
_–7_ * _–27_ = ac = 3 * 63
_–7_ + _–27_ = b = – 34
(3x − 7 ) (3x − 27 )
(3x − 7 ) (3x − 27 )
3
(3x − 7 )(x − 9)
Notice that when a is prime, the extra 3 can be removed from
just one binomial!