How to Factor Trinomials with Understanding!
This video shows the fastest and easiest way I know of for factoring quadratic trinomials. Give it a watch and see if you agree.
Archive for the ‘Practice Problems’ Category
How to Remove (“Unpack”) Algebraic Terms from Parentheses
As you’re probably aware, I’m a big believer in using stories to bring math to life. Especially when you’re teaching tricky concepts, using a story can be the “magic switch” that flicks on the light of understanding. Armed with story-based understanding, students can recall how to perform difficult math processes. And since people naturally like stories and tend to recall them, skills based on story-based understanding really stick in the mind. I’ve seen this over and over in my tutoring.
The kind of story I’m talking about uses an extended-metaphor, and this way of teaching is particularly helpful when you’re teaching algebra. Ask yourself: what would you rather have? Students scratching their heads (or tearing out their hair) to grasp a process taught as a collection of abstract steps? Or students grasping a story and quickly seeing how it guides them in doing the math? I think the answer is probably pretty clear. So with this benefit in mind, let’s explore another story that teaches a critical algebraic skill: the skill of “unpacking” terms locked inside parentheses.
To get the picture, first imagine that each set of parentheses, weirdly or not, represents a corrugated cardboard box, the kind that moving companies use to pack up your possessions. Extending this concept, the terms inside parentheses represent the items you pack when you move your goodies from one house to another. Finally, for every set of parentheses (the box), imagine that you’ve hired either a good moving company or a bad moving company. (You can use a good company for one box and a bad company for a different “box” — it changes.) How can you tell whether the moving company is good or bad? Just look at the sign to the left of the parentheses. If the moving company is GOOD, you’ll see a positive sign to the left of the parentheses. If the moving company is BAD, you’ll spot a negative sign there.
Here’s how this idea looks:
+ ( ) The + sign here means you’ve hired a GOOD moving company for this box of stuff.
– ( ) This – sign means that you’ve hired a BAD moving company to pack up this box of things.
Now let’s put a few “possessions” inside the boxes.
+ (2x – 4) This means a GOOD moving company has packed up your treasured items: the 2x and the – 4.
– (2x – 4) Au contraire! This means that a BAD moving company has packed up the 2x and the – 4.
[Remember, of course, that the term 2x is actually a + 2x. No sign visible means there’s an invisible + sign before the term.]
What difference does it make if the moving company is GOOD or BAD? A big difference! If it’s a GOOD company, it packs your things up WELL. Result: when you unpack your items, they come out exactly the same way in which they went into the box. So since a good moving company packed up your things in the expression: + (2x – 4), when you go to unpack your things, everything will come out exactly as it went in. Here’s a representation of this unpacking process:
+ (2x – 4)
= + 2x – 4
Note that when we take terms out of parentheses, we call this “unpacking” the terms. This works because algebra teachers fairly often describe the process of taking terms out of ( ) as “unpacking” the terms. So here’s a story whose rhetoric matches the rhetoric of the algebraic process. Convenient, is it not?
Now let’s take a look at the opposite situation — what happens when you work with a BAD (boo, hiss!) moving company. In this case, the company does such a bad job that when you unpack your items, each and every item comes out “broken.” In math, we indicate that terms are “broken” by showing that when they come out of the ( ), their signs, + or – signs, are the EXACT OPPOSITE of what they should be. So if a term was packed up as a + term, it would come out as a – term. Vice-versa, if it was packed up as a – term, it would come out as a + term. We show the process of unpacking terms packed by a BAD moving company, as follows:
– (2x – 4)
= – 2x + 4
And that pretty much sums up the entire process. Understanding this story, students will be able to “unpack” terms from parentheses, over and over, with accuracy and understanding.
But since Practice Makes Perfect, here are a few problems to help your kiddos perfect this skill.
PROBLEMS:
“Unpack” these terms by removing the parentheses and writing the terms’ signs correctly:
a) – (5a + 3)
b) + (5a – 3)
c) – (– 3a + 2b – 7)
d) + (– 3a + 2b – 7)
e) 6 + (3a – 2)
f) 6 – (3a – 2)
g) 4a + 6 + (– 9a – 5)
h) 4a + 6 – (– 9a – 5)
ANSWERS:
a) – (5a + 3) = – 5a – 3
b) + (5a – 3) = + 5a – 3
c) – (– 3a + 2b – 7) = + 3a – 2b + 7
d) + (– 3a + 2b – 7) = – 3a + 2b – 7
e) 6 + (3a – 2) = + 3a + 4
f) 6 – (3a – 2) = – 3a + 8
g) 4a + 6 + (– 9a – 5) = – 5a + 1
h) 4a + 6 – (– 9a – 5) = + 13a + 11
Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com Just click the links in the sidebar for more information!
How to Decrease Algebraic Mistakes – Part 5
This is the fifth in a series of posts on how to help students make fewer mistakes in algebra.
Let's Reduce Mistakes in Algebra!
So far I have introduced a form of notation I have developed, the double-slash, which looks like this:
//
and I have described some of the ways that students can use it.
I’ll continue the conversation by showing how this notation can help students combine like terms with greater care.
Fun Math Problem #2
Here is the second in my series of “Fun Math Problems.”
Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.
To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.
The Problem: Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.
The question is: how long were you away?
Image by The Hidaway (Simon) via Flickr
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Answer, Fun Math Problem
Answer to problem about the circular and square pegs and holes.
The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)
For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
2 times the radius squared
This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …
Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366
For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.
This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …
Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854
Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.
Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.
And that is the answer!
FUN MATH PROBLEM — Circling the Square & Vice-Versa
From time to time I will post interesting math problems.
Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit!
I will post the answer to the problems two days later, after people have had time to respond.
To post your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
The problem: Which provides the fuller fit? Putting a circular peg in a square hole, or putting a square peg in a circular hole? To get credit, show all work, and justify your answer by expressing each “fit” as a percent.
A few term-clarifications, to help you do this correctly:
a) By “fit,” I mean the ratio of the smaller shape to the larger shape, expressed as a percent. For
example, if a ratio is 4 to 5, that would represent a “fit” of 80 percent.
b) For the circular peg in the square hole, assume that the diameter of the circle equals the side of the
square. For the square peg in a circular hole, assume that the diameter of the circle equals the diagonal of the square.
c) By “fuller fit,” I mean the larger of the two ratios.
Have fun!
How to tell if a Number is Divisible by 8
I’ve explained a number of divisibility rules lately, offering tricks to tell if numbers are divisible by 2, 3, 4, 5, 6 and 7.
There is also a trick for divisibility by 8, and that’s what I’d like to explain in this post.
Essentially the trick for 8 is a lot like the trick for 4. If you’d like to refresh your memory on how that trick works, just go here. (more…)
(Divisibility) Practice Makes Perfect
As the saying goes, practice makes perfect.
And boy is that true in math! Of the standard school subjects, math requires the most practice, if you want to excel at it.
That being the case, this strikes me as a great time to practice the divisibility tricks we’ve just learned.
There are many skill areas where divisibility tricks are useful — solving proportions, factoring polynomials, multiplying fractions — but one of the most obvious is the critical skill of reducing fractions.
So now I’m offering you a chance to practice your divisibility skills for 2, 3, 4, 5 and 6. We will save the trick for 7 till we have a few more tricks “up our sleeves.”
For the following problems, answer these four questions:
1) Which of these numbers — 2, 3, 4, 5 or 6 — divides evenly into the numerator (NM)?
2) Do the same for the denominator (DNM).
3) Then choose the largest number that divides into both NM and DNM. For these problems, this number will be the GCF.
4) Finally, reduce the fraction by dividing both NM and DNM by this number.
Here’s an example that shows what you’d write:
ex) 24/42
1) NM: 2, 3, 4, 6
2) DNM: 2, 3, 6
3) GCF = 6
4) Answer: 4/7
NOW TRY THESE PROBLEMS:
a) 20/24
b) 25/40
c) 18/48
d) 26/60
e) 21/72
f) 30/85
g) 36/66
h) 56/92
i) 84/102
j) 99/141
ANSWERS:
a) 20/24
1) NM: 2, 4, 5
2) DNM: 2, 3, 4, 6
3) GCF = 4
4) Answer: 5/6
b) 25/40
1) NM: 5
2) DNM: 2, 4, 5
3) GCF = 5
4) Answer: 5/8
c) 18/48
1) NM: 2, 3, 6
2) DNM: 2, 3, 4, 6
3) GCF = 6
4) Answer: 3/8
d) 26/60
1) NM: 2
2) DNM: 2, 3, 4, 5, 6
3) GCF = 2
4) Answer: 13/30
e) 21/72
1) NM: 3
2) DNM: 2, 3, 4, 6
3) GCF = 3
4) Answer: 7/24
f) 30/85
1) NM: 2, 3, 5, 6
2) DNM: 5
3) GCF = 5
4) Answer: 6/17
g) 36/66
1) NM: 2, 3, 4, 6
2) DNM: 2, 3, 6
3) GCF = 6
4) Answer: 6/11
h) 56/92
1) NM: 2, 4
2) DNM: 2, 4
3) GCF = 4
4) Answer: 14/23
i) 84/102
1) NM: 2, 3, 4, 6
2) DNM: 2, 3, 6
3) GCF = 6
4) Answer: 14/17
j) 99/141
1) NM: 3
2) DNM: 3
3) GCF = 3
4) Answer: 33/47
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