Download Factoring ax + bx + c Using the strategy – SPLITTING THE MIDDLE

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Objective: Factoring ax + bx + c
Using the strategy – SPLITTING THE MIDDLE TERM
3) Student/peer
1) Teacher Instruction
problem
Algorithm
a) Make sure it is in standard order
Multiply a*c (We already know how
to arrange in standard order and to
identify a and c)
2) Teacher example problem
Factor 3 x 2 17 x 6
a *c
18
Big Note: What if a=1?
b)
factor a*c in such a way that the factors
add to b (bx=b1x+b2x)
(Note: this is a good place to review the XFactor strategy)
c)
d)
If this step can not be done, the
quadratic is not factorable.
rewrite the expression, splitting the
original middle term. Use the two
factors that you chose in the previous
step
group the 4 terms and factor out
common factors
Factors of -18 could be:
18 and -1 or -18 and 1
9 and -2 or 9 and -2
6 and -3 or -6 and 3.
The two factors with the sum of 17 would be -18 and 1
3x2 -18x + 1x -6
3x2 -18x + 1x -6
Or
(3x2 -18x)+ (1x -6)
So
e)
Use the distributive property again to
write the expression as a product of two
binomials… You have factored it!!
3x(x-6) + 1(x-6)
Since I have common factors (x-6)
it factors to
(3x+1)(x-6)
Rewrite the problem to show the
answer in the factored form
f)
Check:
If I check by multiplying the
factors, I get
(3x+1)(x-6) = 3x2 -17x -6
Summary:
Factor By Split MT.doc
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Factoring trinomial quadratics, ax + bx + c
“splitting the middle term” or “a*c” method
(Algorithm & Linkage justification)
1. Form the product “a*c”
2. Find a pair of numbers b1 and b2 whose product is also “a*c” and whose sum is “b” (if you can’t
find such numbers, it can’t be factored)
3. Split the middle term using b1 and b2; that is, express the term bx as b1x + b2x
4. Now you have 4 terms and you will be factoring by grouping pairs of terms
5. Factor out the common group factor in each pair of terms
6. Check by multiplying. Notice the means-extremes products should be your “split”
----------------------------------------------------------------------------------------------------------------------------------It’s interesting that one of the first steps we teach in factoring is to check for common factors, but we
sometimes lose that step in explaining ways to factor trinomial quadratics.
Remember as we introduced the distributive property?
It may have been when we explained that 2x + 3x could be written as 5x , because
(2+3)x OR x(2+3) allows us to combine like terms (in this case constants), 2 and 3.
And, in the reverse order, that 5x could also be written as 2x + 3x. It didn’t seem like earth shattering news
for such simple examples, but sometimes students missed the concept completely. Maybe because it was too
easy.
Then, we used the distributive property for factoring.
e.g. x2 + 2x became x(x+2), 6x+12 to 6(x+2) ,
-3x+6 to -3(x-2), and so on.
Example 1:
x2 + 2x + 3x + 6 could be grouped, (x2 + 2x) + (3x + 6) and then factored, using the distributive property
on each piece, as x(x+2) + 3(x+2).
Hopefully we notice that there are still common factors, (x+2) that could be factored.
(x+2)(x+3) OR (x+3)(x+2).
----------------------------------------------------------------------------------------------------------------------------------Example 2: Did we show the patterns?
x2 + 3x + 4x + 12 in place of x2 + 7x + 12; AND
x2 + 7x + 12
in place of x2 + 3x + 4x + 12 .
The middle term 7x can be “split” into two terms (Inverse of adding or combining like terms.)
Yes, teachers ALWAYS do problems showing how to combine like terms, but do we show possible ways to
combine terms to achieve a given sum and product?
----------------------------------------------------------------------------------------------------------------------------------Example 3:
x2 + 5x + 4x + 20 in place of x2 + 9x + 20
Why do this? We know that when we teach factoring and it is seen as x2 + 9x + 20 , we ALWAYS say, “we
are looking for factors of 20 whose sum is 9. Then we often jump to trial and error and list binomial factors
(x + )(x + ), and maybe do the product thing and “try” 1 and 20, or 2 and 10, or 4 and 5.
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If we are so lucky (or good) enough to choose 5 and 4, we place them in the parentheses in the right place
and multiply the binomials.
(x + 5 )(x + 4 ) = x2 + 5x + 4x + 20 and simplify (combine like terms) to x2 + 9x + 20 .
It’s pretty interesting (evident) that when the correct factors are chosen AND written into the expression as
x2 + 5x + 4x + 20 , the middle term has been “split” into two parts that add to 9 and multiply to 20,
Now, to finish the example,
x2 + 5x + 4x + 20 is grouped as underlined, or in parentheses (x2 + 5x) + (4x + 20)
then factored using the distributive property again,
x(x+5) + 4(x+5). AND, factoring the common (x+5) we get (x+5)(x+4). Yeh!!!
What about the kid who mixes up the 4 and the 5 and writes x2 + 4x + 5x + 20 ?
He then gets x(x+4) + 5(x+4), then (x+4)(x+5). Cool, huh?
----------------------------------------------------------------------------------------------------------------------------------Will this always work?
When you CAN split the middle term in such a way that the product of the two “like term” coefficients is
the last term (constant), the trinomial IS factorable.
What’s more, if the middle term CAN NOT be split it in such a way that the product of the two “like term”
coefficients is the last term (constant), the trinomial is NOT factorable. (This also works for difference of
squares.)
----------------------------------------------------------------------------------------------------------------------------------Examples 4 – 7: (to show importance of signs)
x2 + 8x +12
x2 -x -12
x2 + 6x +2x +12
x2 +3x -4x -12
x2 + 10x +24
x2 + 6x +4x +24
x2 - 10x +24
x2 - 6x -4x +24
Yes, kids can often “just do it,” but keep following the connections to see patterns and skill development.
----------------------------------------------------------------------------------------------------------------------------------Example 8:
How about 2x2 + 11x +12? What’s different? Look at the method sometimes referred to as the “ac”
method, which is just an extension of the method describe above.
Looking back at the simple examples, when a = 1, we factored c.
Now we factor a*c, or in this case 2*12 or 24, and look for factors of 24 that add to 11.
It is easy to show that 3 + 8 = 11, so 3x + 8x = 11x;
And we can write
2x2 + 11x +12
as
2x2 + 3x + 8x +12,
Then group (associate) 2x2 + 3x + 8x +12
Factor each group x(2x+3) + 4(2x+3)
Which factors to (x+4)(2x+3)
So, 2x2 + 11x +12 = (x+4)(2x+3)
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More examples? Variations?
Example 9:
12x2 - 11x -15
a*c = 12 * (-15) = -180
Factors of -180
. . . . .9 & 20, specifically 9 and -20
12x2 - 11x -15 can be written as
12x2 – 20x + 9x -15
Group 12x2 – 20x + 9x -15
(12x2 – 20x) + (9x -15)
4x(3x-5) + 3(3x-5)
(4x+3)(3x-5)
So,
OR
OR
OR
12x2 + 9x - 20x -15. Take your pick!
12x2 + 9x - 20x -15
(12x2 + 9x) - (20x -15)
3x(4x+3) -5(4x+3)
(3x-5)(4x+3)
Therefore 12x2 - 11x -15 = (4x+3)(3x-5)
----------------------------------------------------------------------------------------------------------------------------------Next example with slight variation:
Example 10:
4x2 + 4x -24
Note that there is a common factor that could/should be
factored out of each term first.
4(x2 + x -6) and it fits into our plan for factoring
Example 11:
6x2 + 2x – 20
Again, there is a common factor 2, so it becomes
2(3x2 + x – 10)
and we just keep the 2 throughout the problem
ac = 3 * (-10) = -30
Factors of -20
. . . . .5 & 6, specifically -5 and 6
So, 2(3x2 + x – 10) can be written as
2(3x2 + 6x – 5x - 10)
OR
2(3x2 - 5x + 6x - 10) . Take your pick!
2
Group 2(3x + 6x – 5x - 10)
OR
2(3x2 - 5x + 6x - 10)
2(3x2 + 6x) – (5x - 10)
2(3x2 - 5x) + (6x - 10)
2[3x(x+2) - 5(x + 2)
2[x(3x – 5) + 2(3x - 5)
2(3x – 5)(x + 2)
OR
2(x + 2)(3x - 5)
----------------------------------------------------------------------------------------------------------------------------------Can it be used to identify trinomials that do not factor?
Example 12:
Does 3x2 - 20x + 7 factor? (3)(7) = 21; factors of 21 are only 1 and 21 or 3 & 7; same sign, so there are
no factors of 21 that combine to be 20. Therefore, 3x2 - 20x + 7 is not factorable.
----------------------------------------------------------------------------------------------------------------------------------How about difference of squares? Trinomial squares?
Example 13:
x2 – 64. Same plan: factor -64 so the factors combine to 0. (8 and -8)
It is easily shown that x2 + 8x – 8x– 64 factors to x(x+8) -8(x+8) and then (x-8)(x+8)
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Connections/linkage are numerous:
Factoring integers,
Distributive property and factoring,
Sum of factors,
Product of factors,
Writing math expressions for the words we use to teach factoring (factors that multiply to be “c” and
combine to “b”,
May save time from trial and error,
Finds use/practice for distributive property that is important in future math applications,
Stresses “factors that add to ---“,
Stresses listing factors,
Aids in determining if a trinomial is factorable,
Ties factoring with multiplying (means-extremes)
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x2
5 x 2 x 10
3
x2
8 x 16
5
Examples to practice factoring quadratics
“splitting the middle term” or “a*c” method
Factor or solve by factoring, if possible
2
x2 7 x 6
4
3x 2
10 x 2 13 x 3
6
8 x 2 12 x 3
7
2x2
8
x 2 12 x 36
9
2 x 2 11x 6
10
4 x2
11
4x2
12
3 x 2 11x 4
0
14
3x 2
0
? x 15
64
0
4x 2
9
0
Be careful
13
8x2
22 x 5
0
9 x 12
Be careful
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