Download Strategic Factoring (Advanced)

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FACTORING AX 2 + BX + C WITH A = 1 (20 MINUTES)
Strategic Factoring
Lesson Plan
Developed by CSSMA Staff
Drafted November 2015
Note: In this lesson plan, we outline general things that the teacher can say to his/her class in order to
teach the relevant material effectively. However, depending on teaching style and other factors, teachers
should feel free to deviate from script and adapt the lesson plan to fit their own style and classroom demands.
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Factoring ax 2 + bx + c with a = 1 (20 minutes)
You might factor simple trinomials using algebra tiles. Now, let’s move away from algebra tiles, and directly
use algebra. This is often better as algebra tiles take up a lot of room to draw and sometimes are very
tedious to write out.
Ask: Factor x 2 − 6x + 8.
Possible answer: the factors are x − 2 and x − 4 (demonstrate on board).
Ask: Factor x 2 − 7x − 8.
Possible answer: the factors are x − 8 and x + 1 (demonstrate on board).
Ask: You may have noticed that in the previous example, you had to add and subtract extra tiles to make the
factoring work. Sometimes this can get really tedious, so let’s think of another way to factor. Let’s return to
our previous example of x 2 − 6x + 8. We figured out that the factors are x − 2 and x − 4. Now, notice that
the two numbers we get after the x is −2 and −4. What’s the value of −2 + (−4) ?
Possible answer: −6
Ask: Now, what’s the value of −2 · (−4) ?
Possible answer: 8
Ask: Do you notice a −6 and a 8 in the original equation?
Possible answer: Yes!
Ask: Let’s try it with another quadratic we already factored. Take x 2 − 7x − 8, which we know factors into x − 8 and x + 1. What’s the value of −8 + 1?
Possible answer: −7
Ask: Now, what’s the value of −8 · 1?
Possible answer: −8
Ask: Do you notice a −7 and a −8 in the original equation?
Possible answer: Yes!
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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FACTORING AX 2 + BX + C WITH GCD( A, B, C) , 1 (10 MINUTES)
Ask: Okay, let’s do one final example to see whether this is a coincidence or not. Now, factor x 2 + 3x − 10
and do the same thing as we just did above: multiple and add the numbers after the x in each term.
Possible answer: the factors are x + 5 and x − 2 (demonstrate on board). 5 + (−2) = 3 and 5 · (−2) = −10
and both are in the original equation.
Ask: So, now, if you want to factor x 2 + 2016x + 2015, what should you look for?
Possible answer: Two numbers that add up to 2016 and multiply to 2015.
Ask: What if I flip the sign in the middle, so we have x 2 − 2016x + 2015, what do we look for now?
Possible answer: Two numbers that add up to -2016 and multiply to 2015.
Ask: Correct! In general, if we have the algebraic equation x 2 + ax + b where a and b are some numbers,
what should we look for?
Possible answer: Two numbers that add up to a and multiply to b.
To do this, we usually make a table. For example, if I wanted to factor x 2 − 6x + 8, I could make a
table of two values which multiply to 8:
Value 1
1
2
4
8
−1
−2
−4
−8
Value 2
8
4
2
1
−8
−4
−2
−1
And now, we look for numbers in the same row which add up to −6. It’s easy to find that −4 and −2 works.
So our factors are x − 2 and x − 4. This is the general technique for trying to factor something that looks
like x 2 + ax + b.
Say: Now, do the first three problems on your handout. In the homework, you’ll get more practice
with ideas like this. You’ll even get to show why this always work!
After approximately 10 minutes, starting covering some of the answers to problems 1-3 on the Student
Handout. You do not need to cover the answers to all of the problems. There are more problems on the
Student Handout than is feasibly coverable in the given time frame, but more advanced students should still
be able to finish all of the problems in time.
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Factoring ax 2 + bx + c with gcd(a, b, c) , 1 (10 minutes)
Ask: How do you factor 4x 2 + 2x ?
Students should know by now how to factor common factors out of a binomial. If this is unclear, use
the distributive property (i.e. a(b + c) = ab + ac to show why this is true). A possible answer to the above
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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FACTORING AX 2 + BX + C WITH A , 1 (20 MINUTES)
question is 2x(2x + 1) .
Ask: Now, what if you are factoring a trinomial like 2x 2 + 6x + 4? What suggestions do you have to
factor this?
Possible suggestions: Split the trinomial into (2x+ )(x+ ) or factor out the common factor of 2.
Note that the greatest common factor of all three terms in the trinomial above is 2. x isn’t in the third
term. So you could factor it into 2(x 2 + 3x + 2) . Are we done?
Not yet! x 2 + 3x + 2 can still be factored. Go ahead and figure it out.
Now, let’s factor some harder stuff.
Ask: Let’s try factoring 4(x + 3) + a(x + 3) . What does both parts of this expression have in common?
Possible answer: x + 3 is in both parts of the expression.
Ask: Therefore, x + 3 is the what of both parts of both sides?
Possible answer: Greatest Common Factor.
Ask: So, using the same strategy as before, can you finish up the factoring?
Possible answer: (4 + a)(x + 3) .
Students can now try problems 4-8 on their handout.
After approximately 10 minutes, starting covering some of the answers to problems 4-8 on the Student
Handout. You do not need to cover the answers to all of the problems. There are more problems on the
Student Handout than is feasibly coverable in the given time frame, but more advanced students should still
be able to finish all of the problems in time.
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Factoring ax 2 + bx + c with a , 1 (20 minutes)
Say: Let’s start with an example using algebra tiles:
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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FACTORING AX 2 + BX + C WITH A , 1 (20 MINUTES)
This one isn’t so bad. After playing around with the tiles, student should get 2x + 1 and x + 3 as the factors.
Say: However, as you probably noticed, algebra tiles are extremely inefficient with trinomials with larger
coefficients. Take 4x 2 − 16x + 15 for example. How do you factor this?
Students may suggest looking for numbers that add up to −16 and multiply to 15. Show them that
(x − 15)(x − 1) doesn’t work.
Say: So (x − 15)(x − 1) doesn’t work. When we expand that out, we get x 2 − 16x + 15 and we are
missing the 4 in front of the x 2 term. Now, let’s think about this for a second, in the final factored form, I’m
going to have something that looks like (___ x + ___)(___ x + ___) where those blanks are either positive or
negative numbers. Now, we know that (___ x + ___)(___ x + ___) = 4x 2 − 16x + 15. When you expand the
left hand side, what creates the 4x 2 term?
Possible answer: the two ___ x terms. Explain by distributive property why no other terms can give you a
term of degree 2.
Ask: Similarly, When you expand the left hand side, what creates the term 15?
Possible answer: the two ___ terms. Explain by distributive property why no other terms can give you a
constant.
Say: So, to factor this expression, we can find two numbers that multiply to 4, and put those two numbers in
front of the x terms. For example, our factored form can look like (2x + ___)(2x + ___) because 2 × 2 = 4.
To fill in the last two blanks, we find two numbers that multiply to 15. Here, both writing a table or guess
and checking are valid methods. Try to finish up the problem from here. Possible answer: the two ___
terms are −3 and −5. So the final factored form is (2x − 3)(2x − 5) . Tell students that one could have started
with (4x + ___)(x + ___) and would have gotten stuck: this is where the guess and checking part comes in.
Say: I will show this technique with another example: 12x 2 − 11x + 2.
Step 1: I write 12x 2 − 11x + 2 = (___ x + ___)(___ x + ___) where the blanks are some numbers to be filled
in.
Step 2: I find two numbers that multiply to 2 and put them in the two ___ blanks in the above expression.
What are some pairs of numbers that multiply to 2? This isn’t a trick question. (Wait for a student to
say: "1 and 2 or −1 and −2") We’ll write those pairs down on the side and see which one of these is
correct later.
Step 3: I find two numbers that multiply to 12. So I make a list of pairs of numbers that multiply to 12. I
have (1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1).
Step 4: Now comes the guessing and checking part. I try some pairs of numbers to get a −11x term. But
before we dive into doing that, can we narrow down which pair for the ___ blanks is correct? Is it −1
and −2 or 1 and 2. Think about it. We have a −11x term. Can we get a negative x term with a bunch
of positive coefficients?
Step 5: Guessing and checking....(try some combinations)...and we have it! (3x − 2)(4x − 1) is our winner!
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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HARDER TECHNIQUES (20 MINUTES)
Students can now try problems 9-11 on their handout.
After approximately 10 minutes, starting covering some of the answers to problems 9-11 on the Student Handout. You do not need to cover the answers to all of the problems. There are more problems on the
Student Handout than is feasibly coverable in the given time frame, but more advanced students should still
be able to finish all of the problems in time.
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Harder Techniques (20 minutes)
Now, let’s try the rest of the problems on your handout. First, let’s look at when our x is weird and...kind of
disgusting.
Problem 12: Factor 6(x + 1) 2 + (x + 1) − 2.
Step 1: Think of x + 1 as y , and rewrite the original expression in terms of y .
Step 2: You should have 6y 2 + 6 − 2. Now, factor this normally.
Step 3: Factoring should give (2y − 1)(3y + 2) . Now, plug back y = x + 1 into the expression.
Step 4: You should have (2(x + 1) − 1)(3(x + 1) + 2) . This simplifies to (2x + 1)(3x + 5) . That’s our
answer.
Students can now try problem 13.
Say: Let’s now look at problem 14. Now, it’s not even a quadratic...what the? How do we do this?
Say: Hmmm, we notice that the powers of x ’s are both even, so maybe if we do a substitution like in
the last problem, like let y equal to some power of x , we might be able to simplify this thing down to a
quadratic...any suggestions?
Possible answer: Let y = x 2 ; let y = x 4 (show why this doesn’t work and y = x 2 works by using
exponent laws).
Problem 14: Factor x 4 − 4x 2 + 3.
Step 1: Think of x 2 as y , and rewrite the original expression in terms of y .
Step 2: You should have x 4 − 4x 2 + 3 = (x 2 ) 2 − 4x 2 + 3 = y 2 − 4y + 3. Now, factor this normally.
Step 3: Factoring should give (y − 3)(y − 1) . Now, plug back y = x 2 into the expression.
Step 4: You should have (x 2 − 3)(x 2 − 1) . That’s our answer.
Students can now try problem 15.
Say: Finally, let’s look at problem 16. This one is tough.
Problem 16: Factor (x 2 − 3x + 2)(x 2 − 3x − 4) − (x 2 − 6x + 8) .
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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HARDER TECHNIQUES (20 MINUTES)
Step 1: Well, it doesn’t seem like we have many choices other than to factor each of the three quadratics
separately. Let’s do that first then.
Step 2: You should have (x − 1)(x − 2)(x − 4)(x + 1) − (x − 2)(x − 4) as the fully factored form.
Step 3: Do you now see that (x − 2)(x − 4) is the common factor of both terms in the previous expression?
(x − 1)(x − 2)(x − 4)(x + 1) − (x − 2)(x − 4)
|
{z
} |
{z
}
has a (x-2)(x-4) in it
has a (x-2)(x-4) in it
Step 4: So, treating (x − 2)(x − 4) like a common factor, we pull it out to get:
(x − 2)(x − 4)[(x − 1)(x + 1) − 1]
Step 5: Finally, expanding the stuff inside the square brackets give
(x − 2)(x − 4)(x 2 − 2)
which gives us our final answer!!!
Students can now try problem 17.
Give students their Homework assignment. Teaching ends here, and if there’s still more time, allow
students to work on the homework assignment for the remaining duration of class. Allow students to discuss
problems with each other as collaborative learning can be very effective!
At the beginning of the next class (or in two classes’ time, depending on how comfortable the students
seem to be with the material at the beginning of the next class—a generally weaker class might need more
time than an Honors class, for example), collect the homework and discuss some of the problems on it. Right
after discussing some of the problems, administer the Quiz at the end of this lesson plan. The suggested
time frame is 20 minutes, but depending on how comfortable your class seem to be with the material, this
number can be raised or lowered per your will.
The homework is made to be quite challenging, so we actually do not expect every student to solve every problem. The total number of marks in this homework is 30 points, but teachers should feel free to
lower than number appropriately to reflect the climate of the classroom. The teacher may even tell his/her
students to skip some of the hardest problems ahead of time (or make them bonus questions). For example,
problem 3 is quite challenging, so that can be made a bonus question.
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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STUDENT HANDOUT
Student Handout
1. Factor x 2 + 9x − 70 by looking for two numbers that add up to 9 and multiply to −70.
2. Factor x 2 − 2x − 120. Try to cleverly guess the two numbers that add up to −2 and multiply to −120
without making a full sized table.
3. Alphonse is factoring x 2 − 60x + 800. He claims that the expression factors into (x − 10)(x − 80) .
Beryl looks at it and immediately tell Alphonse that he’s wrong. How did Beryl figure out so quickly?
4. Factor 500x 2 + 1000x − 7500.
5. Factor 6x 2 − 24x + 18. What about −6x 2 + 24x − 18?
6. Factor 4y 2 (3y − 1) + 3y − 1.
7. Factor x 2 (y 2 + 2y − 8) + x(y 2 + 2y − 8) . This one is a bit more tricky!!!
8. There are a number of ways to write (6x + 12)(8x − 4) as a factored form. For example, one
of them is (3x + 6)(16x − 8) (why is (3x + 6)(16x − 8) = (6x + 12)(8x − 4) ?). How many ways
are there to write (6x+12)(8x−4) in a factored form so that there isn’t a number in front of the brackets?
9. Factor 8x 2 − 10x − 3 and 6x 2 − 7x − 3.
10. (a) Let’s think about factoring for a moment. Suppose we want to factor x 2 + 8x + 15 (don’t do it
yet). We have x 2 + 8x + 15 = (x + ___)(x + ___) . Without actually doing the factoring, are the
blanks both positive numbers, both negative numbers, or one positive and one negative?
(b) Suppose we want to factor x 2 −8x +15 (don’t do it yet). We have x 2 −8x +15 = (x + ___)(x + ___) .
Without actually doing the factoring, are the blanks both positive numbers, both negative numbers,
or one positive and one negative?
(c) Suppose we want to factor x 2 +8x −20 (don’t do it yet). We have x 2 +8x −20 = (x + ___)(x + ___) .
Without actually doing the factoring, are the blanks both positive numbers, both negative numbers,
or one positive and one negative?
(d) Suppose we want to factor x 2 −8x +20 (don’t do it yet). We have x 2 −8x +20 = (x + ___)(x + ___) .
Without actually doing the factoring, are the blanks both positive numbers, both negative numbers,
or one positive and one negative?
(e) Does having a different coefficient in front of the x 2 term make a difference? Consider 2x 2 −9x+10
and x 2 − 9x + 8 for example.
(f) Describe the general rule about the signs of the blanks when you factor ax 2 + bx + c into
(___ x + ___)(___ x + ___) . If a is positive, assume that both of the blanks in front of the x term
are positive. What if a is negative?
11. Factor 6x 2 − 26x + 8. (Hint: gcd(6, 26, 8) = 2.)
12. Example Problem: Factor 6(x + 1) 2 + (x + 1) − 2.
13. Factor 4(2x 2 + x − 6) 2 − 19(2x 2 + x − 6) + 12. This one requires one more step than the previous
example problem.
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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STUDENT HANDOUT
14. Example Problem: Factor x 4 − 4x 2 + 3.
15. Factor x 6 + 6x 3 + 5.
16. Let’s combine the last two techniques: factor (2x − 1) 4 + 2(2x − 1) 2 − 8.
17. Example Problem: Factor (x 2 − 3x + 2)(x 2 − 3x − 4) − (x 2 − 6x + 8) .
18. Factor (x 2 + 3x + 2)(x 2 + 7x + 12) + (x 2 + 5x + 6) .
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.
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HOMEWORK
Homework
Name: ___________________________________
/30
Today’s homework involves both a bit of review as well as you doing a bit of discovering. Please write your
solutions on separate sheets of paper. Make sure to show all of your work and reasoning. There will
be a quiz on factoring next class.
1. (a) (1 mark) Factor x 2 − 13x − 48.
(b) (1 mark) Factor x 2 + 7x − 60.
(c) (2 marks) Factor 2x 2 + 3x − 5.
(d) (2 marks) Factor 20x 2 − 11x − 45.
(e) (3 marks) Factor 24x 3 + 22x 2 + 3x .
2. Let’s first see why the following statement is true:
Theorem 6.1. If you can factor x 2 + ax + b (a and b are integers) into x + r and x + s (r and s are
two other integers), then r + s = a and r s = b.
(a) (1 mark) Since x 2 + ax + b factors into x + r and x + s, we must have x 2 + ax + b = (x + r)(x + s) ,
right? So, expand (x + r)(x + s) and use your answer from (a) to show that ax + b = (r + s)x + r s.
(b) (1 mark) Since x is a variable, you can plug in x = 0 into both sides of the expression. Some
terms should cancel out, so what’s left?
(c) (1 mark) You should have gotten an equality in the previous problem. Plug that equality back
into ax + b = (r + s)x + r s. After cancellations, what do you get?
(d) (1 mark) Now, assume that x , 0, so divide through by x on both sides. What do you get? You
should now be able to conclude that r + s = a and r s = b.
3. I want to factor x 2 + ax + b where a and b are integers into (x + r)(x + s) where r and s are integers.
(a) (2 marks) If b = 18 and r = 3, determine the value of a.
(b) (3 marks) If b = 12, find all possible values of a.
(c) (2 marks) If a = 2, how many possible values for b are there if b must be smaller than 20 and
greater than −20?
4. (3 marks) Factor (x + 4) 4 − (2x + 8) 2 . (Hint: there are a lot of common factors!)
5. (3 marks) Factor (x 2 − 2x) 2 + 3(2x − x 2 ) − 4.
6. How would you compute the following without the use of a calculator, quickly and efficiently:
(a) (2 marks) 19952 + 11 × 1995 + 30
(b) (2 marks) 1998 × 1997 + 3 × 1998 + 2 × 1997 (you can almost factor this one...what do you need
to add to make it factorable?)
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QUIZ: NO CALCULATORS ALLOWED
Quiz: No Calculators Allowed
Name: ___________________________________
Date: ____________________________________
1. (1 mark) Factor x 2 − x − 110.
2. (2 marks) Factor 3y 2 − 15y − 18.
3. (2 marks) Factor 6x 2 − 13x + 6.
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QUIZ: NO CALCULATORS ALLOWED
4. (2 marks) Factor y 5 + 7y 3 − 18y .
5. (3 marks) I have the expression 2x 2 − ax − 15 which can be factored into the form (2x + r)(x + s)
where r and s are integers. If a is a prime number, find the possible value(s) of a.
Copyright Canadian Secondary School Mathematics Association (CSSMA). 2015-2016. All rights reserved.