### 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.

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?

- Clocks for Math Geeks (geeksaresexy.net)

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!

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!

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…)

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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