Bay Area Tutoring

Factoring ax² + bx + c, part 2

By Tutor GuyNo Comments

In my last post, I showed you a procedure for how to factor a quadratic when the leading coefficient is not equal to 1. Here I refine that method slightly with a short cut that makes your work even quicker. However, this method is a bit more difficult—if you are not really good at mental calculations, you will need to practice this a few times to get it down. To compare methods, I’ll use the same quadratic we factored in the last post.

Problem: Factor $12x^2+4x-5$

Our first step is the same as in the previous method—we need to find two numbers that multiply together to give $60$ (that’s $12*-5$ ) and add together to give $+4$ . Of course, those numbers are the same as they were in the last post: $+10$ and $-6$ .
……
Now here’s the shortcut. Divide both of these numbers by the leading coefficient, $12$ . That gives two fractions:
…..
….. $\dfrac{10}{12} \;\; \text{and } \dfrac{-6}{12}$
…..
Next, simplify the fractions if possible. In our example, we get
…..
….. $\dfrac{5}{6} \;\; \text{and } \dfrac{-1}{2}$
…..
In each fraction, the denominator is the coefficient of the x term in our factor and the numerator is the constant term. So our two factors must be $6x+5$ and $2x-1$ . See the diagram below:

Feb 09

Factoring ax² + bx + c

By Tutor GuyNo Comments

You probably have a lot of experience factoring quadratics of the form $x^2+bx+c$ . (That is, where $a$ , the coefficient on the $x^2$ term is equal to $1$ .) But what do you do when $a$ isn’t $1$ ? For example, let’s factor $12x^2+4x-5$ . You may have been taught a trial-and-error technique where you look for all the numbers that multiply to give $12x^2$ and combine them with all the numbers that multiply to give $5$ and test each one to see which gives the correct middle term. But there’s a simple algorithm that takes all the guesswork out. I’ll show you the procedure with an example.

Problem: Factor $12x^2+4x-5$

Similar to the case where $a=1$ , look for two numbers that multiply together to give $a \cdot c$ (in our example, $12 \cdot -5=-60$ ) and add together to give $b$ (i.e., $+4$ in our example). [Actually, you can see that this is just like the $a=1$ case, because there you were looking for two numbers that multiply together to give $c$ , but when $a=1$ , $a \cdot c=c$ …] For our example, those numbers are $+10$ and $-6 \\$ .
…..
Bring down the first term without changing it:
…..
….. $12x^2$
…..
Write the linear term as a sum or difference using the two numbers you found in step A.:
…..
….. $12x^2$ $+10x-6x$
…..
(Don’t worry about which term comes first. As it turns out, you can write these two terms in either order.)
…..
Bring down the last term without changing it:
…..
….. $12x^2$ $+10x-6x$ $- \; 5$
…..
Now comes the magic. Look at the first two terms only from step D (cover up the last two terms if that helps you) and factor them:
…..
….. $2x$ $\!\!(6x+5)$
…..
Now look at the last two terms in step D (again, cover the first two terms if that helps you) and factor them. And this is very important: if the $x$ term has a minus sign on it, always factor out the minus sign:
…..
….. $2x$ $\!\! (6x+5)$ $- \; 1$ $\!\! (6x+5)$
…..
(Note that if nothing factors out of the last two terms, write a $1$ (you can always factor $1$ out of any term).
…..
Are the two expressions in the parentheses the same? They have to be. If they’re not, you’ve done something wrong. That’s one of the factors in your answer. The other factor is made up of the numbers outside the parentheses:
…..
….. $(2x-1)$ $\!\! (6x+5)$

Tada! I’ll leave it to you to show that if you had written the linear terms in step C in the reverse order, you’d get the same answer.

Feb 02

The addition rule and multiplication rule in probability

By Tutor GuyNo Comments

Two of the many formulas you learn in statistics are referred to as the Multiplication Rule and the Addition Rule. How do you figure out when to use each one? This is actually a pretty easy one. The multiplication rule is used to determine the probability of two events both happening, one after the other. The key word to look for in a problem is ‘and’.

Example 1: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven first AND a heart second? Because you want the probability of both events occurring, you use the multiplication rule.

The addition rule is used to find the probability that either one of two events occurs. The key word here is “or’.

Example 2: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven on the first draw OR a heart on the second draw? Because you want the probability of either event occurring, you use the addition rule.

How do the rules work? The multiplication rule comes in two forms. If the two events are independent, that is, if the probability of the second event does not change when the first event occurs, then the formula is simple:

$P(A \cap B) = P(A) \cdot P(B)$

If the probability of the second event depends on the first event occurring, the formula is modified slightly to show this:

$P(A \cap B) = P(A) \cdot P(B|A)$

You can see from either formula why this is called the multiplication rule. The addition rule is just slightly more complicated.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

You can see from this formula why it is called the addition rule. Let’s solve both problems.

Example 1: The two events are independent, so we use the first version of the rule:

$P(A \cap B) = P(A) \cdot P(B)=\dfrac{4}{52} \cdot \dfrac{13}{52}= \dfrac{1}{52}$

Example 2: (Note that we use the result from Example 1 in solving this problem.)

$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\dfrac{4}{52}+ \dfrac{13}{52}- \dfrac{1}{52}= \dfrac{16}{52}= \dfrac{4}{13}$

Jan 26

Finding the inverse of a 3X3 matrix

By Tutor GuyNo Comments

What’s the easiest way to find the inverse of a 3×3 matrix? Use your graphing calculator, of course! But if you need to find the inverse without a calculator, here’s a method that will give you the solution with the least amount of trouble. We’ll demonstrate with an example.

Find the inverse of the following matrix:

$\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{bmatrix}$

The first step in finding the inverse is to calculate the determinant of the matrix. The easiest way to calculate a 3×3 determinant is to write the matrix out, and append the first two columns at the end:

$\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{bmatrix} \begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6\\ \end{matrix}$

From here, you need to find the six different products along each of the diagonals shown:

Add the blue products together and add the red products together, then subtract the red total from the blue total: $(50 + 96 + 84) - (105 + 48 + 80) = -3$ . This is the determinant. By the way, if the determinant is $0$ , stop. Your matrix does not have an inverse.

Next, you need to find the elements of the inverse matrix. Here’s a clever trick that will help you do that. Start by writing the transpose of the original matrix. This is done by changing all the rows into columns:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}$

Then write the first two columns over on the right and the first two rows over again on the bottom. Your array should look like this:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix} \begin{matrix} 1 & 2 \\ 4 & 5 \\ 7 & 8 \end{matrix} \\ \begin{matrix} \text{ }1 & 2 & 3 & \,\, 1 & 2 \\ \text{ }4 & 5 & 6 & \,\, 4 & 5 \end{matrix}$

The next step is a little tricky to explain, though once you’ve done it, it’s pretty easy to figure out. For each of the nine positions in the matrix, you find the value of the determinant of the 2×2 array that is just below it and to the right. [One way to think of this is that each position in the matrix is the upper-left number of a 3×3 array. If you mentally delete the first row and column of that array, you have a 2×2 array left. This is the array for which you find a determinant. I demonstrate this below with the $1$ and the $6$ from the array above.

Do this process for all nine elements of the matrix and you should end up with this matrix:

$\begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}$

Finally, divide this matrix by the determinant you found above. The result will be the inverse of the original matrix:

$\dfrac{1}{-3} \begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}= \begin{bmatrix} ^{-2} \! /_3 & ^{-2} \! /_3 & 1 \\ ^{-4} \! /_3 & ^{11} \! /_3 & -2 \\ 1 & -2 & 1 \end{bmatrix}$

I’ll leave it to you to verify that this is indeed the inverse of the original matrix. For any 3×3 matrix that has an inverse, this method will calculate it for you correctly every time.

Jan 19

Completing the square

By Tutor GuyNo Comments

Are you an expert at completing the square? This is an important skill that is not nearly as difficult as a lot of students think it is. If you master just a couple of rules, you will be able to complete squares like a pro.

Let’s start with an explanation. Since we complete the square in algebra and not in geometry, exactly where is the square we are completing? This confuses many students and is probably the reason why they are frightened unnecessarily by the process. But it’s actually pretty simple. Let’s say we have the following expression:

$(x-3)^2$

This is obviously the square of $(x-3)$ , and so we call it a perfect square. If we expand the expression into $x^2-6x+9$ , we recognize that it is still a perfect square. Any trinomial of the form $ax^2+bx+c$ is a perfect square if it is the square of a binomial expression.

When we have an expression such as $x^2-6x$ , we look for a number we can add to make it a perfect square of a binomial. In this case, adding $9$ to $x^2-6x$ to get $x^2-6x+9$ gives us the square of $x-3$ . Therefore we have completed the square. That’s all the expression means!

Completing the square is actually quite easy in most cases. Because $(x+a)^2=x^2+2ax+a^2$ , anytime we have an expression of the form $x^2+2ax$ , all we need to do is add $a^2$ . You can do this in two quick steps:

Take the coefficient on the $x$ term and divide it by $2$ .
Square this number and add it to the end.

For example:

$x^2+8x$ : Divide the $8$ by $2$ (you get $4$ ) and square it (you get $16$ ), so the completed square is $x^2+8x \; +$ $16$
$x^2-12x$ : Divide the $-12$ by $2$ (you get $-6$ ) and square it (you get $36$ ), so the completed square is $x^2-12x \; +$ $36$
$x^2+9x$ : Divide the $9$ by $2$ (you get $\frac{9}{2}$ ) and square it (you get $\frac{81}{4}$ ), so the completed square is $x^2+9x \; +$ $\frac{81}{4}$
$x^2-13x$ : Divide the $-13$ by $2$ (you get $\frac{-13}{2}$ ) and square it (you get $\frac{169}{4}$ ), so the completed square is $x^2-13x \; +$ $\frac{169}{4}$

Next you write the factor as a square. The factor is always of the form $(x+a)^2$ or $(x-a)^2$ where the value of $a$ is the number that you squared. For the four examples above:

$x^2+8x \; +$ $16$ $= (x$ $+ \; 4$ $)^2$
$x^2-12x \; +$ $36$ $= (x$ $- \; 6$ $)^2$
$x^2+9x \; +$ $\frac{81}{4}$ $= (x$ $+ \; \frac{9}{2}$ $)^2$
$x^2-13x \; +$ $\frac{169}{4}$ $= (x$ $- \; \frac{13}{2}$ $)^2$

What if your expression starts with $ax^2$ where $a$ is not equal to $1$ ? Then you have to factor the $a$ out of both terms before you complete the steps above. It can appear very complicated, but the process is the same every time. Here are some examples:

$2x^2+8x=2(x^2+4x) \rightarrow 2(x^2+4x+4)=2(x+2)^2$
$4x^2-12x=4(x^2-3x) \rightarrow 4(x^2-3x+ \frac{9}{4}) = 4(x- \frac{3}{2})^2$
$2x^2+9x=2(x^2+ \frac{9}{2}x) \rightarrow 2(x^2+ \frac{9}{2}x+ \frac{81}{16}) = 2(x+ \frac{9}{4})^2$
$3x^2-13x=3(x^2- \frac{13}{3}x) \rightarrow 3(x^2- \frac{13}{3}x+ \frac{169}{36})=3(x- \frac{13}{6})^2$

Dec 29

Sum( and Seq( commands on your calculator

By Tutor GuyNo Comments

Can you find the sum of the following series?

$\displaystyle \sum_{i=1}^{15}\dfrac{2n+1}{3n-2}$

This is neither an arithmetic nor a geometric series, so you don’t have a formula for it. This would be a tedious problem to do by hand. Fortunately, your graphing calculator can do these problems quickly and efficiently.

There are two functions you need to use on your calculator. The seq( command creates a sequence of terms based on a rule that you give. The sum( command adds together the terms in a sequence. Both functions are found on the LIST menu on your calculator. The seq( command is on the OPS submenu and the sum( command is on the MATH submenu.

To sum a series, you combine the two commands. If you have the new operating system on your calculator, it will prompt you for the entries when you select the seq( command. If you have the old operating system, you need to know the syntax for the command. The syntax for the series above is:

sum(seq((2x + 1)/(3x – 2),x,1,15))

Note that the seq( command has four parameters in the parentheses. From left to right, these are 1) the rule for the nth term of the sequence; 2) the variable name; 3) the first value of the variable; and 4) the final value of the variable. Now all you need to do is type this in to your calculator and let it do the crunching:

Dec 22

The “parent” functions

By Tutor GuyNo Comments

Do you know your parent functions? I’m surprised every year when some of my students don’t know how to graph some of the elementary functions they’ve learned in class. Your chances for success in precalc and calculus are significantly better if you memorize the shapes of these ten functions. No excuses, no shortcuts. Just learn them.

y = xⁿ (Power functions)

y = |x| (absolute value function)

y = ⌊x⌋ (Greatest integer function)

y = √x

y = 1/x

y= b^x (exponential function – growth and decay)

y = log x

y = sin x and y = cos x

y = tan x

Dec 15

Calculating probabilities as a fraction

By Tutor GuyNo Comments

The first day you are asked to find probabilities, the questions are relatively simple:

What is the probability of rolling a 2 or a 3 on a standard die? [Ans: 2/6 = 1/3]
What is the probability of drawing a five from a standard deck of cards? [Ans: 4/52 = 1/13]

But very quickly, the problems get a lot more complicated:

What is the probability of flipping a coin ten times and getting exactly 8 heads?
What is the probability of being dealt a full house (three of a kind plus a pair) in a five card poker hand?
A class has 14 boys and 12 girls. If the teacher randomly creates a new seating chart, what is the probability that the six students in the front row comprise exactly four boys and two girls?

How do you attack these problems? The best way is to think of probability as a fraction. The numerator counts all the ways the specified event can occur, and the denominator counts all the ways any event can occur:

Your event
Any event

Then you use combinatorics to count all the possibilities on the top and on the bottom. Let’s see how we can use this method to answer the three questions posed above:

$st-0401-frac$

(Note that this question could also be treated as a binomial probability and solved that way.)

$st-0402-frac$

Dec 08

Factoring sums and differences of fifth powers (and higher!) when the power is odd

By Tutor GuyNo Comments

Occasionally, you are required to factor a polynomial that is in the form of a sum or difference of two power terms (i.e., xⁿ– yⁿ). In my last post, I showed you a simple procedure you can learn to help you factor the binomial if the value of the power is 4 or greater and the power is even. [If you are reading this post, you should already be familiar with factoring the difference of two squares and the difference of two cubes. Check this website for posts on those procedures.] In this post, I show you a procedure (more complicated than the others, I admit) that you can use to factor these binomials when the power is odd.

First, let’s see how this works on some examples:

Do you see the pattern? When factoring aⁿ– bⁿ or aⁿ+ bⁿ (for n odd), there is always a linear term a ± b, where the plus or minus sign is the same as in the original binomial. The remaining factor is a little more complicated, but it does have a simple pattern. Each term consists of the first term in the binomial (with its exponent decreasing from n – 1 to 0) and the second term in the binomial with its exponent increasing from 0 to n – 1. In the first example above, the x term decreased from x⁴ to x⁰. The 2 term increased from 2⁰ to 2⁴. If you started with the difference of two powers, all of the signs in the second factor are plus signs. If you started with the sum of the two powers, the first sign in the second factor is a minus sign and the signs alternate after that.

Dec 01

Factoring differences of fourth powers (and higher!) when the power is even

By Tutor GuyNo Comments

Occasionally, you are required to factor a polynomial that is in the form of a difference of two power terms (i.e., xⁿ – yⁿ). There is a simple procedure you can learn to help you factor the polynomial if the value of the power is 4 or greater and the power is even. [If you are reading this post, you should already be familiar with factoring the difference of two squares and the difference of two cubes. Check this website for posts on those procedures.]

First, let’s see how this works on some examples:

You can see that whenever the power is even, you treat the binomial as the difference of two squares and find its factors. If the factors are still the sum and difference of perfect squares, you repeat the process with the term that is the difference of two squares. If the factors are now the sum and difference of perfect cubes, you follow the method for factoring the sum or difference of two cubes. If the powers are sums and differences of odd powers higher than 3, there is another procedure to factor them. You can find another post on this website that shows that procedure.

Archive

For the Algebra 2 category

Factoring ax² + bx + c, part 2

Factoring ax² + bx + c

The addition rule and multiplication rule in probability

Finding the inverse of a 3X3 matrix

Completing the square

Sum( and Seq( commands on your calculator

The “parent” functions

Calculating probabilities as a fraction

Factoring sums and differences of fifth powers (and higher!) when the power is odd

Factoring differences of fourth powers (and higher!) when the power is even

Tutor Guy Homework Tips