Math: Factoring: Difference of Squares

The math tutor recommends a little light factoring on this beautiful Sunday morning….

Last post I discussed common factor, which we will be using in concert with difference of squares. Difference of squares factors x2 – 36 into (x + 6)(x – 6). By the foil method you can confirm:

F: x*x=x2

O: x*-6=-6x

I: 6*x=6x

L: 6*-6=-36

Displaying the terms in a row we get

x2 – 6x + 6x – 36 = x2 – 36

Example: Factor x2 – 49

Solution: We notice that the square root of 49 is 7. Therefore we write

x2 – 49 = (x + 7)(x – 7)

Difference of squares is easy to spot and factor if it’s plain. However, it may be “hidden” by a common factor:

Example: Factor 2x3 – 50x

Solution: We notice that 50x isn’t square rootable. However, we also notice that 2x can be taken out front as a common factor:

2x3 – 50x = 2x(x2 – 25)

Now, we’re getting somewhere: we follow with

2x(x2 – 25) = 2x(x + 5)(x – 5)

Removing the common factor of 2x allowed us to apply the difference of squares technique.

Have a nice day:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Factoring: Common Factor

As a math tutor, you notice the importance of this technique.

Factoring means breaking a number or expression into a product.  For instance, we’ll factor 45:


In earlier posts I’ve mentioned prime factorization:


Now we’ll look at factorization of polynomials using common factor.
Example: factor -2x6 + 8x5-12x2

Solution: With the common factor method, we look for the expression that divides into all the terms, then write it out front. What remains in the brackets is each term divided by the common factor.

In this case we notice that 2x2 divides into all the terms. Therefore, we “take it out front”. Actually, we take out -2x2 because whenever the lead term is negative, you take out the negative with the common factor. Inside the brackets we write each term divided by -2x2:

-2x2(x4 -4x3+6)

Common factoring doesn’t have to be as complicated as the example above.  Consider the following:

3x – 15 factors to 3(x – 5)

Working with polynomials, factoring is constantly used.  There are at least five factoring techniques, of which common factor is the first.  I’ll discuss the other techniques in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Direct Proportionality

As a math tutor, you’ll likely introduce this concept.  It’s used even more in physics and chemistry.

So often in my university science courses I’d read “the mass is directly proportional to the volume” or “the distance is directly proportional to the time”, etc.  Science people love direct proportionality because predicting the result of a given input is so easy.

Direct proportionality means that if you double the input, the output will also double.  If y is directly proportional to x, it follows that


In the above equation, is called the constant of proportionality. Once you know k, you can find the result of any input.

Example:  The distance travelled by a long haul train is directly proportional to the time traveled.  The train travels 600 km in 9 hours.

a)  Find the equation to model this situation.
b)  How far will the train travel in 12 hours?

Solution:  y is always the “output”, while x is the “input”.  Some people like to use different letters in order to reflect the actual wording of the question.  In that case:

(1)   \begin{equation*}d=kt\end{equation*}

where, of course, d stands for distance, t for time.

To find k, we use the idea that the train travels 600 km in 9 hours:

(2)   \begin{equation*}600=k(9)\end{equation*}

Dividing both sides by 9, we get

(3)   \begin{equation*}\frac{600}{9}\ =k\end{equation*}

(4)   \begin{equation*}66.7=k\end{equation*}

We know now that the equation to model the train’s travel is

(5)   \begin{equation*}d=66.7t\end{equation*}

To predict the train’s distance over 12 hours, we simply put 12 in for t:

(6)   \begin{equation*}d=66.7(12)\end{equation*}


(7)   \begin{equation*}d=800\end{equation*}

We conclude that over 12 hours, the train will travel 800 km.

More will be said about direct proportionality in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Biology: Energy and ATP

Tutoring biology, you need to be aware of the connection between ATP and energy.

In a factory or a mill, there likely is a power plant where fuel is burnt en masse. The energy is captured there (often in the form of electricity), then channeled to the other locations as needed.

The human body, though, releases and consumes energy in a different way.  Each cell receives fuel (glucose), burns it internally, then captures the released energy by using it to synthesize a high energy chemical bond.  That bond can be broken at will when energy is needed.

ADP is adenosine diphosphate (adenosine bonded to two phosphate groups). When the cell burns glucose (in its mitochondria), the energy released is used to bond another phosphate to the ADP, so it becomes ATP (adenosine triphosphate).

When the cell needs energy, it breaks an ATP into an ADP and a phosphate.  The breaking of that bond releases energy the cell can use to power any life process.  Later, when the cell burns more glucose, it will use the energy to bond the ADP and the phosphate back into ATP.

The burning of one glucose molecule produces 36 or 38 ATP.

Hope this helps:)

Source:  Inquiry into Life, Eleventh Edition, Sylvia S. Mader.  McGraw-Hill:  2006.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Finding Square Root or Cube Root from Prime Factorization

The math tutor continues to appreciate prime factorization for all it yields.

Let’s imagine you need to determine the square root of a number without a calculator. This challenge is part of the curriculum for local high school students.

Example:  Determine if each number can be square rooted (to a whole number).  If so, find its square root.

a)  540
b)  576

Before tackling the above problem, let’s dissect a number we know to be a perfect square.

Example:  Confirm, by prime factorization, that the square root of 900 is 30.

Solution:  We recall that a prime number is one that cannot be divided into smaller numbers, then break 900 down into primes:


Rearranging, we get


We notice 900 can be broken into two identical groupings like so:

900 = (2x3x5)(2x3x5)

Therefore, the square root of 900 is 2x3x5=30

We now know what to seek:  if a number is square rootable, its prime factorization can be organized into two equal groups.  The square root is simply the product of one of the groups.

Back to our example:

Determine the whole number square root (if it exists) of the following:

a) 540
b) 576


a)  First we break 540 into primes:


With only one 5 in the prime factorization, we can’t separate it into two equal groups. 540 doesn’t have a whole number square root.

b)  2 and 4 both go into 576.  Without a calculator, you either do it mentally or else use long division.  To get started, just break it in half:


Since we have only multiplication here, we can add and rearrange brackets at will. However, with mixed operations we wouldn’t be able to do so:)

Rearranging, we get


Clearly, the prime factorization of 576 is separable into two equal groupings of 2x3x4. 2x3x4 = 24, so the square root of 576 is 24.

If its prime factorization can be separated into three equal groupings, the number is a perfect cube:

Example:  Confirm that 9261 is a perfect cube.

Solution:  We’ll break this one down using short division.  Since 9+2+6+1=18, we know 9 divides into it:

Since 1+0+2+9=12, we know 3 divides into it (because 3 divides into 12).

Since 3+4+3=10 (which 3 doesn’t divide into), and 343 doesn’t end in 5 or 0, the next number to try is 7:

So we see that we can break down 9261 as follows:


Rearranging, we separate the prime factorization into three equal groupings:


Therefore, 9261 is a perfect cube with cube root=3×7=21.  The cube root of 9261 is 21.

Once again, short division was used to break down 9261.  For more explanation about that very handy technique, please check future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Introduction to Polynomials

As a math tutor, you deal with polynomials half the weeks of the year.

A polynomial is an expression in which the variables can have only positive, whole-number exponents.  Examples of polynomials are 3x7-12x5+1 or  -2x – 12.

In a polynomial, terms are separated by plus or minus signs.  Therefore, 7x – 12 has two terms, whereas  -13x7yz2 has only one.

Polynomials are often named by how many terms they have, as follows:

7x2                   One term: monomial

3x – 12             Two terms: binomial

-4x4 + 3x -5     Three terms: trinomial

A polynomial with more than three terms is just called a polynomial.

Consider the following monomial:


-11 is called the coefficient.  x is called the variable.  5 is the exponent.

The degree of a polynomial is the highest exponent found in one of its terms.  For example, 5x3 has degree 3. The trinomial

x7 + 3x3 – 12

has degree 7.

The constant term of a polynomial is the term with no variable attached. In 3x – 12, the constant term is -12.

There are two facts about polynomials that might be a little surprising:

Fact 1:

A constant term has degree zero.  Reason: x0= 1 by definition. The result:

3x0 = 3(1) = 3.

We just write 3, but the degree of the term is still zero.

Fact 2:

If the coefficient is not written, it is 1.  Reason: x5 = 1x5.

There are a couple of finer points, but the above is good for a start.  Much more will be said about polynomials in future posts.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.