September 22, 2019

## Common Core Math 2 – Geometry Unit 1 Vocabulary

As I begin to help students at , a tutoring center in Apex, NC prepare for Common Core Math 2 / Geometry, I have decided to include some of the material online to make it available to all students who are beginning this course. The first thing that a student needs to know about Geometry is that one must learn all of the vocabulary and theorems in order to be successful. There is no other way to be successful in this class. Therefore, I will start with a list of the beginning vocabulary that one should learn and focus on as they begin to study for this course.

Line – Most students know what a line is but it is important to note that a line has arrows in both directions, note it is different from a line segment that terminates with end points.

Line Segment – A part of a line that terminates with a point at either end.

Ray – A mix between a line and a line segment, one end has a point, the other end an arrow.

Plane – Think of a desktop that continues infinitely in all directions.

Co-linear – Points that are co-linear would be points that fall on the same line.

Midpoint – This is the exact middle of a line segment, it divides something into two equal halves. The formula for calculating a midpoint doesn’t need to be memorized if you just remember that it is the average of the x-values and the average of the y-values.

Acute angle – A “cute” small angle that is greater than 0 degrees but less than 90 degrees.

Right angle – An angle that is exactly 90 degrees. We put a box in the corner of the angle to show it is right. In Geometry, we cannot assume an angle is right just by looking at it, if we don’t have proof or see the “box” we cannot assume it is 90 degrees even if it “looks” right.

Obtuse angle – An “obese” angle or fat angle, one that is greater than 90 but less than 180.

Straight angle – A angle that forms a line and measures 180 degrees.

Complementary – Two angles who sum to 90 degrees.

Supplementary – Two angles who sum to 180 degrees.

Adjacent angles – Angles that are next to each other.

Linear Pair – Two adjacent angles that are supplementary.

Bisect – Something that cuts into two equal pieces such as an angle bisector would cut the angle into two equal pieces.

Vertical Angles – Angles opposite each other (often form an X) – vertical angles are complementary.

Perpendicular Bisector – A line that bisects another line by hitting it at a right angle and cutting it into two equal pieces.

Distance between two points – Memorize distance formula or learn how to use the Pythagorean Theorem to find the distance between any two ordered pairs.

Perimeter – Distance around an object.

Circumference – The distance around a circle: C = PI X Diameter.

Area – The space inside a shape.

Radius – The distance from the center of the circle to the one end of the circle. Radius is half the diameter.

Diameter – The distance from one side of a circle to the other going through the center of the circle. Diameter is twice the radius.

Addition Property of Equality: When you add the same number to both sides of an equation, it doesn’t effect the equality of the equation.

Subtraction Property of Equality: When you subtract the same number from both sides of an equation, it doesn’t effect the equality of the equation.

Multiplication Property of Equality: When you multiply the same number to both sides of an equation, it doesn’t effect the equality of the equation.

Division Property of Equality: When you divide the same number to both sides of an equation, it doesn’t effect the equality of the equation.

* The above four properties are what you do when you solve an Algebraic Equation such as: 2x – 5 = 11 (add 5 to both sides: Addition property of Equality, then divide both sides by 2, Division property of equality).

Substitution Property: If two things are equal you may substitute one for the other: measure of angle 1 = measure of angle 2 and measure angle 2 + measure of angle 3 = 180, since measure of angle 1 = measure of angle 2, I can SUBSTITUTE the measure of angle 1 into my other equation making it: measure of angle 1 + measure of angle 3 = 180.

Transitive Property: (Remember as the 3 piece property) If a = b and b = c then a = c. If the measure of angle 1 = measure of angle 2 and the measure of angle 2 = measure of angle 3, then I know that the measure of angle 1 must also equal the measure of angle 1.

Reflexive Property: (Think Reflection): a=a Something always equals itself. It may seem obvious but is needed in proofs.

Angle Addition Postulate: If you have a big angle divided into two pieces the two pieces add together to equal the total angle.

Corresponding Angles: When two parallel lines are cut by a transversal, the angles that correspond with each other are congruent.

Alternate Interior (and Exterior) Angles: When two parallel lines are cut by a transversal, the angles that are on opposite sides of the transversal line but are both inside (or both outside) the parallel lines) are congruent.

Same Side Interior (and Exterior) Angles: When two parallel lines are cut by a transversal, the angles that are on the same side of the transversal line but are both inside (or both outside) the parallel lines are supplementary.

Remembering Congruence of Angles: Vertical Angles, Corresponding Angles, Alternate Interior (Exterior) Angles
Remembering Supplementary Angles: Linear Pairs, Same Side Interior (Exterior) Angles

These are the vocabulary words one should first learn and also be able to apply to concrete problems and proof situations as they start CCM 2 / Geometry! Good Luck!

So, to all of you who are considering grade skipping – know that others may not understand but as long as you know that it is the best decision for your child as a whole, their is a lot of research to support your decision.  My school district also says it isn’t a reversible decision either – if you try it and it doesn’t seem to be working out, you can always choose to go back to your child’s same age peers.

## Should Partial Credit Be Awarded on Math Tests?

This is a debatable subject. Math teachers seem to be on one side or the other. When asked for reasoning, I hear things such as, “No partial credit should be given because the real world doesn’t allow for things to be wrong.” Other teachers are very busy and don’t have the time to look at a student’s work in the detail needed to figure out where their mistake was or what their thinking was, so they could correctly award partial credit; it is much easier to just grade it as 100% right or 100% wrong. On the other side of the coin, teachers who award partial credit encourage students to “show their work” and want to encourage students for getting conceptual parts of the problem correct and not penalize them for making one tiny mistake in a multi-step problem that demonstrates that they have actually learned what they were being taught. We are all human after all.

So – what is the correct approach? Should partial credit be awarded? If so, how much should be awarded? When should it be awarded? Do teachers have the “right” to choose the 100% right / 100% wrong approach? Is it fair for some teachers to grade students this way, hence awarding a B or C to a student that might actually have a good grasp of the content when another teacher who gives partial credit would give that same student a grade 10 points higher – and hence the “unlucky” students who get the non-partial credit teachers look like they understand less (when in fact they don’t) than another student who happens to have a teacher who awards partial credit?

Are all math teachers flawless? If they were not to use a calculator at all, would they never transpose a number or accidentally make a mistake? Of course not, all teachers (myself included) have been corrected by students when we occasionally make a mistake during our lessons. Yet, we are willing to subtract 8-10 points off a test grade if the student does the same?

What if the problem is testing a very difficult concept and the student gets all the concept correct, showing they clearly understood everything taught to them but they accidentally transpose a number or maybe made a silly arithmetic mistake or even lost a negative sign in all the written work required as they were focusing on the difficult concept. Are we then to reward them with no credit when in fact they clearly learned what we were trying to teach them?

Here is a quote from Brian Boley, “Avoid the “partial credit” trap when teaching middle school and high school students. Someday you may drive over a bridge which one of your students designed. Do you expect him to have calculated the loads correctly or should he get “partial credit” for getting a close answer? And all because you taught him that using the right equation was worth 90% of the problem — and adding 2 + 5 = 8 was only 10% off.”  His argument is sound, right?  Who would want to be on that bridge?  Yet, is that what partial credit is promoting?  Would we give credit for 2+5 = 8?  Of course not, that is wrong.  The entire concept that is being taught is wrong and hence no partial credit should be award in those cases.  When the student misses the concept, they do in fact lose all credit for the problem.  If they do a math problem in Algebra and have no understanding – just a few random ideas – that is not a time to offer partial credit.  We are talking about giving credit to the student who made a careless error but who clearly understood what they were doing.  Remember, we are not building a bridge, if we were, we wouldn’t have a new student learning something for the first time doing the math for it – that is not how the real world works – school is a time for learning.  A “close answer” does not equal credit, what equals credit is a demonstration of the concept being tested or a partial demonstration of that concept that shows you got 1/2 the concept and you missed 1/2 the concept so we will award you credit for the 1/2 of the concept you got correct and take away credit for the 1/2 of the part of the concept that you still need to learn.  It is just a way to break up the scoring of a problem with multiple steps into multiple scoring which is a fair and reasonable thing to do.  So, when you hear arguments like Brian’s, don’t immediately think – yeah, I don’t want to be on that bridge – don’t worry, brand new Algebra students or middle school students don’t build bridges.

What is our goal in teaching mathematics? Don’t we want people to stop saying, “I am not any good in math.” Well, we will just continue to perpetuate this problem by not rewarding students with partial credit especially when it is obvious that they grasped the concept being taught and the mistake was elsewhere! Why don’t math teachers care about our students’ perceptions of mathematics? How can you choose to be a math teacher when you don’t care enough to make students want to feel good about math. Now, don’t get me wrong, I am not a proponent of teachers who give grades for undeserved work! I met a woman who wanted her students to feel positive about math so she gave everyone a B or higher – no matter what they did. That won’t help them either. They must earn their grades but if you give them positive feedback, encouragement (which includes acknowledging their efforts and what they have learned and accomplished with partial credit), they will respond with a better self-image about mathematics which in return will improve their efforts, attitude, study habits, and hence their grades.

I also don’t agree that life only allows for Right and Wrong answers. If that were the case, we would all be in a lot of trouble. We are human, we make mistakes, it is a great thing for kids to learn that we acknowledge that we all make mistakes and don’t expect perfection and that the world does not expect perfection. Even working a job, people will make mistakes, if you do, you figure out where your mistake is, you communicate with others, you realized the solution is not working so you rework the problem and find your own mistake, etc. Very few people do everything perfect the first time in the real world. Why would we penalize our kids psychological well-being as well as their future (see grade issue above based on the 100% wrong teachers vs. the partial credit teachers) because they made a small arithmetic mistake even though they correctly integrated this very long function?!

I also think we owe it to them to look at their work and try and find their mistakes or if teachers don’t have the time, get creative. Mark it wrong and let the student come back with a test correction where they show the teacher where their mistake was and offer them partial credit back at that point based on WHY they got the problem wrong. It makes the student go back and find their own mistakes and yet still gives them partial credit.

Award partial credit appropriately. If the mistake was just arithmetic and the concept was Algebra – they lose a little. If they transpose numbers but did the whole problem right with the transposed numbers – they lose almost nothing! If they make a partial Algebra mistake – they lose much more credit, depending on how much of concept they were able to get. For example, if solving an Algebra word problem, if they got the equation right but then had no idea how to solve the equation – they would get half (or more than half as finding the equation is really the hard part) credit – if they got the equation and just made a “mistake” solving the equation but seemed to know the general process, they lose less. So partial credit is not awarded equally. If a student is solving an order of operations problem and they do the order of operations correct but state that 4^2 = 8 instead of 16, they should be awarded a large part of the credit since the problem was testing order of operations but lose some for not knowing how to evaluate an exponent. If they make that same mistake again in future problems but again solve the problem correctly, the amount lost should be minimal since obviously they will continue to make that same mistake but you already took off for it and the teacher should be looking for the main idea of the question, not marking every problem on the test wrong just because the student missed this one concept of how to evaluate an exponent even though they can solve everything else about the other problems correctly.

So, to answer the question – Should partial credit be awarded on a math test? The answer is a resounding YES. I hope this article points to the many reasons why it is important to award partial credit to students on their tests.

Author: Lynne M. Gregorio, Ph.D. in Mathematics Education
Owner: Triangle Education Center and Educator for over 23 years.

## Geometry: Translations, Reflections, Rotations, and Dilations

One unit covered in Geometry deals with the concept of translations, reflections, rotations, and dilations.  This article will serve to summarize some the major points for students studying these topics.

Prerequisite:  The student needs to come into the lesson with some basic understanding of matrices.  Given a shape with points on a coordinate plane they need to be able to write those in matrix form.  It is very simple actually, you take the x-coordinates and make them the first row of your matrix and take the y-coordinates and make them the second row of your matrix.

Example: A quadrilateral with points (-4,-3), (-1,0), (1,-3), and (-3, -5) would be written as the following matrix:

$$\left[ \begin{array}{ c c c c } -4 & -1 & 1 & -3\\ -3 & 0 & -3 & -5 \end{array} \right]$$

Students also need to know the identity matrix when multiplied by a matrix gives back the original matrix.  It is like multiplying a number times 1.  The identity matrix is:

$$\left[ \begin{array}{ c c } 1 & 0 \\ 0 & 1 \end{array} \right]$$

Translations: Translations simply slide your figure around.  It is the easiest to work with since it just involves adding a value to the x-coordinates and a value to the y-coordinates.

Example: Translate the example matrix above by moving it to the RIGHT four and DOWN 1.  This would mean we just add 4 to the top numbers and subtraction 1 from the bottom numbers:

$$\left[ \begin{array}{ c c c c } -4 & -1 & 1 & -3\\ -3 & 0 & -3 & -5 \end{array} \right] + \left[ \begin{array}{ c c c c } 4 & 4 & 4 & 4\\ -1 & -1 & -1 & -1 \end{array} \right]= \left[ \begin{array}{ c c c c } 0 & 2 & 5 & 1\\ -4 & -1 & -4 & -6 \end{array} \right]$$

Dilations: Dilations make an object bigger or smaller.  If the dilation is a number bigger than 1, the object will increase in size; if it is less than 1, it will get smaller.  Dilations require you multiple each number in the given matrix by the dilation value.

Example: Dilate the given quadrilateral by 3.

3 * $$\left[ \begin{array}{ c c c c } -4 & -1 & 1 & -3\\ -3 & 0 & -3 & -5 \end{array} \right] = \left[ \begin{array}{ c c c c } -12 & -3 & 3 & -9\\ -9 & 0 & -9 & -15 \end{array} \right]$$

The “slightly” harder problems involve ROTATION and REFLECTION.

These simple “adjust” the coordinates according to a specific matrix.  Let’s look at some different matrices and see what they do:

Identity: Doesn’t change the value of the matrix.

$$\left[ \begin{array}{ c c } 1 & 0 \\ 0 & 1 \end{array} \right]$$

Matrix 1: Notice this looks just like the identity matrix except it has negative 1’s rather than positive ones.  This means that it will change each sign to its opposite in the matrix.

$$\left[ \begin{array}{ c c } -1 & 0 \\ 0 & -1 \end{array} \right]$$

Matrix 2: This matrix looks like the identity but has a negative only in the top 1.  This means only the top row will change to their opposite signs but the bottom row will stay the same.

$$\left[ \begin{array}{ c c } -1 & 0 \\ 0 & 1 \end{array} \right]$$

Matrix 3: Matrix 3 is similar to Matrix 2 but the negative is on the bottom instead of the top.  This means the bottom row will change to its opposite sign and the top row stays the same.

$$\left[ \begin{array}{ c c } 1 & 0 \\ 0 & -1 \end{array} \right]$$

Matrix 4: This matrix is a little different from the identity.  The 1’s and the 0’s have changed places.  When this happens, the whole row changes places.  Since both are positive, the numbers keep their original signs.

$$\left[ \begin{array}{ c c } 0 & 1 \\ 1 & 0 \end{array} \right]$$

Matrix 5: Can you guess what happens in this matrix?

$$\left[ \begin{array}{ c c } 0 & -1 \\ 1 & 0 \end{array} \right]$$

$$\left[ \begin{array}{ c c } 0 & 1 \\ -1 & 0 \end{array} \right]$$

Matrix 7: And this one?

$$\left[ \begin{array}{ c c } 0 & -1 \\ -1 & 0 \end{array} \right]$$

Matrix 5 – the rows switch places and the top row has opposite signs.

Matrix 6 – the rows switch place and the bottom row has opposite signs.

Matrix 7 – the rows switch places and both rows also change signs.

Each of these matrices are multiplied times the matrix defined by the shape in the problem  Note that the “identity” type matrix always comes first, then the other matrix so that the dimensions match for multiplying.

Here is a summary of when to use each matrix:

Reflection over y = x: use matrix

$$\left[ \begin{array}{ c c } 0 & 1 \\ 1 & 0 \end{array} \right]$$

Reflection over x-axis: use matrix

$$\left[ \begin{array}{ c c } 1 & 0 \\ 0 & -1 \end{array} \right]$$

Reflection over y = -x: use matrix

$$\left[ \begin{array}{ c c } 0 & -1 \\ -1 & 0 \end{array} \right]$$

Reflection over y-axis: use matrix

$$\left[ \begin{array}{ c c } -1 & 0 \\ 0 & 1 \end{array} \right]$$

Rotation of 90 degrees: use matrix

$$\left[ \begin{array}{ c c } 0 & -1 \\ 1 & 0 \end{array} \right]$$

Rotation of 180 degrees (same as rotation over Ho): use matrix

$$\left[ \begin{array}{ c c } -1 & 0 \\ 0 & -1 \end{array} \right]$$

Rotation of 270 degrees: use matrix

$$\left[ \begin{array}{ c c } 0 & 1 \\ -1 & 0 \end{array} \right]$$

Example: Given A(2,5) and B(1, -2) and C(-2,3).

Find a rotation of 270 degrees:

$$\left[ \begin{array}{ c c } 0 & 1 \\ -1 & 0 \end{array} \right] * \left[ \begin{array}{ c c c } -2 & 1 & -2 \\ 5 & -2 & 3 \end{array} \right] = \left[ \begin{array}{ c c c } 5 & -2 & 3\\ -2 & -1 & 2 \end{array} \right]$$

Note:  The two rows switched, then the bottom row switches signs.

Find a reflection over the y-axis:

$$\left[ \begin{array}{ c c } -1 & 0 \\ 0 & 1 \end{array} \right] * \left[ \begin{array}{ c c c } 2 & 1 & -2 \\ 5 & -2 & 3 \end{array} \right] = \left[ \begin{array}{ c c c } -2 & -1 & 2\\ 5 & -2 & 3 \end{array} \right]$$

Note: The rows did not switch, but the signs on the top row changed to their opposites.

To summarize:

Dilations: Multiply matrix through by the amount of the dilation.

Translation:  Adjust each x by the change in the x-axis of the translation and adjust each y by the change in the y-axis  of the translation.

Reflections and Rotations:  Find the corresponding matrix for each reflection or rotation and multiply the matrix by the correct “identity-type” matrix listed above.

Happy Transformations!

www.apex-math.com

Part 2 and 3 of the video can be found under Videos on the homepage.

# Teaching Multiplication Facts

Multiplication facts are part of the North Carolina Standard Course of Study curriculum for the third grade.   Eventually we hope that students will just “know” their facts – in other words, they are memorized but when first learning facts, it is best to teach students strategies for finding facts.  This allows students to always have a fall back plan in case they “forget” the fact and it makes them quicker to learn. The order in which facts should be taught is given below.  The reason for this is that the easier ones get learned first and then they can rely on their “partner fact” (3X4=4X3) for some of the harder facts.

Here are some methods for each fact:

X0 Fact: Anything X0 is 0. This is a very easy fact since students just need to learn that anything times 0 is 0.  Remind them that 0 sets of something is 0.

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X1 Fact: Anything X1 is itself. Again, another very easy fact.  For example,  1 X 7, this means 1 set of 7, which is just 7 items.

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X2 Fact: Circle the number that is not 2, double that number. Since the student is very good at doubling, this is an easy fact.

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X4 Fact: Circle the number that is not 4, double the number and double again.   The student should have doubling down well before starting, so X4 facts will come very easy to them.  The only one that might be difficult is X9 – tell them that they can wait and use the “9’s trick” on that one instead of the 4’s trick if they don’t know 18 + 18 since we didn’t really drill that double.

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X10 Fact: Add 0 to the number that is multiplied by 10.

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X5 Fact: There are two strategies that work here.  Lots of kids like to just count by 5’s because they are good at it.  That takes longer when they are counting by 5 eight times for numbers like 5 X 8.  For these bigger numbers (especially the even ones) they can try this other strategy:   multiply the number times 10 (see 10 fact) and then take half of that number:  so for 5 X 8 we would do 10 X 8 = 80 then take half = 40.

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X11 Fact: Circle the number that is not 11, write the number twice – 11 X 3 = 33.

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X9 Fact: Use the finger trick – hold up all ten fingers, count from the left and bend down the finger of the number you are multiplying by 9. For example, for 9X3, bend down your third finger. Now count the number of fingers on the left side of the bent finger, write that number down – then count the number of fingers on the right side of the bent finger, that is your second digit. With the 9X3 example, you have 2 fingers to the left of the bent finger and 7 fingers on the right side of the bent finger, the answer is therefore 27.  Also show your child all the different patterns in the 9’s.  The sum of the digits of all the 9 facts add to be 9.  The tens place of the 9’s facts is always 1 less than the number you are multiplying by:  if you multiply 9 X 6 then you know that your answer is going to be Fifty – something since 5 is one less than 6.

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X3 Fact: This is where we get the harder ones – students should know the commutative property and therefore only need to learn: 3X3, 3X6, 3X7 and 3X8. There is no good trick for these, tell the child to double the number and add one more. 3X3 = (double 3) + 3 = 6 + 3; 3X7 = (double 7) + 7 = 14 + 7 = 21.  They can also learn to count by 3′s.

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X6 Fact: With the commutative property, you will only need to learn 6X6, 6X7, and 6X8. 6X6=36 and 6X8 can be taught using rhythm (tap them out on the table as you say them).

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X7 Fact: For this group, you will need to learn 7X7 and 7X8. To memorize 7 X 8, I use the visual:  5 6 7 8 these numbers go in order of counting and you can just put = and x in the middle and get your fact 56 = 7 x 8. Help your child see this visual and use it as a cue to help them to remember the fact.  7 X 7 = 49 – sorry, just got to memorize that one.

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X8 Fact: You only need to learn 8X8=64. You can use “bend down touch the floor, eight times eight is 64” or “Skate X Skate = Sticky Floor.”  You can also do “double, double, double if the student can double 3X.  This triple double works really well for 8 X 3 as it is a harder 8 fact but is easy to double 3X.  Double 3 gets to 6, double 6 gets to 12, double 12 gets to 24.

You will want to work on 1 fact at a time, do lots of practice with that fact before moving to the next one. Once you move to the second fact, you should work on that alone and then provide a mix of that with all previously learned facts, and so on.

TABLE of FACTS – shows strategies for each of the different facts:

ANY CHARACTER HERE

You will want to work on 1 fact at a time, do lots of practice with that fact before moving to the next one. Once you move to the second fact, you should work on that alone and then provide a mix of that with all previously learned facts, and so on.  Note that the facts above are written in the order in which they should be taught.  Don’t teach them in numerical order but instead from easiest to hardest so that students can use the commutative property to their advantage and have a feeling of accomplishment.

We are getting ready to launch our Online Multiplication Curriculum. This online program will be interactive and affordable. It will teach your child each strategy mentioned and give them practice with instant feedback using the strategies. We hope to post it on our site soon but for those who can’t wait, feel free to contact us about our pre-release version.

# Teaching Division

In order to learn division, the student must first have a good understanding of multiplication. The child doesn’t need to be perfect but should know the majority of the facts or have a reasonably quick strategy to figure out the answer.

When teaching division, we will be going in steps.

Step 1: Have the student understand the concept of division and be able to solve division problems with manipulatives.

For example: Given 12 pennies. Have the student evenly divide the 12 pennies among 2 people. From that practice writing the division statement 12 / 2 = 6 and have student explain the meaning: “12 pennies divided into 2 groups gives 6 in each group.” Then have the student divide the 12 into 3 groups and repeat the math sentence and the explanation. Next, divide 12 into 4 groups, then 6 groups. Repeat with other numbers such as 16 (divide into 2, 4, and 8 groups) until the student shows mastery of the concept and writing the corresponding math sentence.

Step 2: Have the student understand the concept of a remainder. You will continue with manipulatives in this exercise. Give the child 5 pennies. They have to share the pennies among 2 people. Let them try, if they split the pennies into 2 and 3 then discuss how that isn’t fair. If they divide it 2 and 2 with 1 left over, explain that happens sometimes with division: we don’t have an even amount to divide and therefore get a remainder. Have the student write the problem as: 5 / 2 = 2 with Remainder 1. Continue this process with other numbers: 9 divided by 2. 11 divided by 3, etc.

Step 3: Have the student understand the link between multiplication and division. Go back to your manipulatives and have them show you 3 x 4 = 12 with manipulatives. Remember that multiplication should have been taught as the x means “groups of” so 3 x 4 means 3 groups of 4. Put your 3 groups of 4 equals 12 to the side. Now have them take 12 and divide it 3 groups as you did in step 1. 12 / 3 = 4. Show them that their result is the same as their multiplication piles they made. In the end they have 3 groups of 4 items. Have the student write the fact family: 12 / 3 = 4; 12 / 4 = 3; 3 x 4 = 12; 4 x 3 = 12. Show them how if you read a division problem “backwards” you have a multiplication problem. Also show them that they can think of 12 / 3 as “what times 3 equals 12?” Continue with practice – 1) writing fact families and 2) finding missing factors: 4 X ___ = 16 (After they get the answer, convert to a division problem – 16 / 4 = 4).

Step 4: The next step is to teach the long division process. The key here is that we are focusing on the process, not on learning division – although doing the practice will help reinforce the concepts of division. Print out and cut the division cards, you will need to use these are you teach the process.

Example: 215 / 5.

We actually start with a 3 digit number because it shows the repetitive process. Get out the X5 division card. On a white board, write the problem. The steps to the division process are:

1. Look at the first number, 2, does 5 go into 2? No, 5 is too big.

2. Look at the first 2 numbers together: 21. Looking at the division card, find the number closest to 21 without going over. You see that the number is 20.

3. Write the number to the left (blue number) above the 1 on top. Write the number to the right (red number) below the 21. So, a 4 goes on top and the 20 goes below.

4. Now, subtract 21-20. You get 1. Using an arrow (make sure they use the arrow) bring down the 5 so it is next to the 1, making 15.

5. Repeat process: find a number as close to 15 without going over. We find 15 in the table. The red number (3) goes on top above the 5. The blue number (15) goes below the 15, now subtract. We get 0. Note, there are no more numbers to bring down and since we ended with 0, we have no remainder.

Example: 3426 / 5

1. 5 doesn’t go into 3.

2. Find closest to 34 on chart without going over: 30 (5X6=30).

3. Put 6 on top and 30 under the 34.

4. Subtract, get 4.

5. With arrow, bring down the 2, to get 42.

6. Find the closest to 42 without going over: 40 (5X8=40)

7. Put the 8 on top and the 40 under the 42.

8. Subtract, get 2.

9. With arrow, bring down the 6, to get 26.

10. Find the closest to 26 without going over: 25 (5X5 =25)

11. Put the 5 on top and the 25 under the 26.

12. Subtract, get 1.

13. Since that was our last number, 1 is the remainder.

14. The answer is 685 Remainder 1.

Keep practicing with different divisors until the student can do it independently with the division cards.

Step 5: The student now knows the concept of division and the process of division. Now they need to practice division without the help of the division strips. First give them problems with one of the easy divisors such as 2 or 5 and a 2 digit dividend. Even if it goes in evenly such as 5 into 25, make sure they write out the 25 underneath and show the remainder goes to 0. Give them remainders that go to zeros and ones that don’t so they get practice taking the problem to the end.

Step 6: Once the student is successful with 2 digit dividends with divisors of 2 and 5, have them work with 3 and 4 digit dividends but still use the divisors of 2 and 5.

Step 7: Expand with new divisors one at a time. Do 2 digit dividends first and then expand to 2or 4 digit dividends before moving onto a new divisor.

To view division strips visit:  http://www.apex-math.com/teaching-division-to-children