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Algebra Basics: What Are Functions? - Math Antics

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Understand How to Graph Lines in 10 min (y=mx + b)

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Factoring quadratic expressions: how to walkthrough (video) | Khan Academy

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Can’t get enough of Sal factoring simple quadratics? Here’s a handful of examples just for you!

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Factoring completely with a common factor (video) | Khan Academy

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We can factor quadratics by first pulling out a common factor so the result looks like a(x+b)(x+c).

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Hey, check out Padlet. I think you’ll like it. https://itunes.apple.com/us/app/padlet/id834618886?mt=8

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Make sure it's 2014 and newer version.

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Math Antics | Basic Math Videos and Worksheets

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If You Love Math Explained In A Way You Like And Understand - Then You Will Love TabletClass Math Academy!

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Colleges and employers accept the GED credential

Copyright © 2020 GED® Testing Service LLC. All Rights Reserved.

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Colleges and employers accept the GED credential Copyright © 2020 GED® Testing Service LLC. All Rights Reserved. Your GED ® account is your one-stop shop for passing the GED ® test. It's free to set up, and you'll find study materials, tips, and classes.

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CLASS JOIN LINK

https://quizlet.com/join/DD2ecvqt5

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Join a game of kahoot here. Kahoot! is a free game-based learning platform that makes it fun to learn – any subject, in any language, on any device, for all ages!

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permutation and combination using calculator

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permutation and combination using calculator

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TI 30SX Calculator Tutorial   Probability Functions

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Scientific Notation and the TI-30XS MultiView Calculator

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Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube.

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Most Missed GED® Math Questions - Part 1

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Great site to obtain math cheat sheets

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All students were given a booklet at the beginning of the school year entitled:

Hardin Middle School Math Cheat Sheets.

If the cheat sheet book has been lost, it can be replaced for $5. You can access the entire book, or just sections of the book below.

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All students were given a booklet at the beginning of the school year entitled: Hardin Middle School Math Cheat Sheets. If the cheat sheet book has been lost, it can be replaced for $5. You can access the entire book, or just sections of the book below.

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Visual Fractions: Online Calculators & Tools

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Online references and tools. Fraction and math calculators, converters, crossword solvers and much more!

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Great site for math tutorials

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Algebra is great fun - you get to solve puzzles!

With computer games you play by running, jumping and finding secret things.

With Algebra you play with letters, numbers and symbols, and you also get to find secret things!

And when you learn some of the "tricks" it becomes a fun challenge to work out how to use your skills in solving each question.

The Basics

Exponents

Simplifying

Factoring

Logarithms

Polynomials

Linear Equations

Quadratic Equations

Solving Word Questions

Functions

Sequences and Series

Where to Next?

Other Interesting Topics You May Like

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Algebra is great fun - you get to solve puzzles! With computer games you play by running, jumping and finding secret things. With Algebra you play with letters, numbers and symbols, and you also get to find secret things!

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GED Math - How to Get the Right Answers on the 2020 Test (1)

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Lesson on graphing linear equations Part 2

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System of Equation: In real life examples

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Algebra Basics: What Is Algebra? - Math Antics

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Algebra Basics: Solving Basic Equations Part 1 - Math Antics

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THE USE OF POLYNOMIAL FUNCTIONS IN REAL LIFE

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Polynomials are fit to data points in both regression and interpolation. In regression, a large number of data points is fit with a function, usually a line: y=mx+b. The equation may have more than one "x" (more than one dependent variable), which is called multiple linear

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Dividing Polynomials: Reduction & Long Division [fbt]

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Butterfly method for adding or subtracting fractions

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Butterfly method for adding or subtracting fractions

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worksheets polynomials (add and subtract)

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worksheets polynomials (add and subtract)

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Difficult Trinomial Factoring Using the Criss-cross Method; Example 3

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Factoring Trinomials (ax^2 + bx + c) using Diamond & XBox Method

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Purplemath

A "quadratic" is a polynomial that looks like "", where "a", "b", and "c" are just numbers (and either of b and c, but never a, might be zero).

For the easy case of factoring quadratic polynomials, we will need to find two numbers that will multiply to be equal the constant term c, and will also add up to equal b, the coefficient on the linear -term in the middle. Why?

Quadratics are degree-two polynomials. When you were learning how to multiply polynomials together, they probably started you out by multiplying two binomials together (and they probably referred to this process as "FOILing").

MathHelp.com

This quadratic has a leading coefficien of 1, so this is the simple case of factoring. To get started, I need to find factors of c = +6 that add up to b = +5. I have two options, because 6 factors as the product of 2 and 3, or as the product of 1 and 6.

Now, because I'm multiplying to a positive six, then my factors have to have the same sign; they must both be positive or else they must both be negative, because that's how negatives work. Because I'm adding to a positive five, then both factors must be positive.

I'll check the sums of the pairs of potential factors, to see which works:

Since I need my factors to sum to plus-five, then I'll be using the factors +2 and +3.

From back when I learned how to multiply polynomials, I know that they got this quadratic by multiplying two binomials. Because the leading coefficient is just 1, I know that the leading coefficient of each of those binomials must also have been just 1. This means that the product started out looking something like this:

At the ends of each parenthetical go the numbers that multiply to +6 and add to +5. This means that I can finish my factoring by plugging those numbers into my parentheticals, in either order:

The leading term is just 1, so this is the simple case of factoring. The constant term is 6, which can be written as the product of 2 and 3 or of 1 and 6, just as in the previous exercise. But the coefficient on the middle term is different this time. Instead of a +5, I've got a +7.

The sign on the constant term is the same as before (namely, a "plus"), so I'll still need to "plus" factors. But the sum (that is, the middle-term's coefficient) is different; it's now a 7. Whereas 2 + 3 = 5 worked for the previous quadratic, +2 and +3 are not the numbers I need in this case. On the other hand, 1 + 6 = +7, so I'll use +1 and +6 for my factorization. And I'll also go straight from having drawn this conclusion to writing down my final answer:

The constant term (created by multiplication) is +6, so my factors will be either both "plus" or else both "minus". But the middle coefficient this time is "minus". Since I'm adding to a "minus", (namely, to -5), then both factors must be "minus".

When the middle term's coefficient had been a "plus" five, I'd used factors +2 and +3. Now that my middle term's coefficient is "minus", I will use -2 and -3:

If c is "plus", then the factors will be either both "plus" or else both "minus".

If b is "plus", then the factors are both "plus".

If b is "minus", then the factors are both "minus".

In either case, look for factors that add to b.

If c is "minus", then the factors will be of alternating signs; that is, one will be "plus" and one will be "minus".

If b is "plus", then the larger of the two factors is "plus".

If b is "minus", then the larger of the two factors is "minus".

In either case, look for factors that subtract to b; that is, factors which are b units apart.

The leading coeficient is 1, so this is a simple-factoring quadratic. I am multiplying to a "plus" six, so the factors will be either both "plus" or else both "minus". Looking at the middle term, I see that I am adding to a "minus" seven, so my factors will both be "minus".

The (negative) factors of +6 that add up to 7 are -1 and -6, so I will use -1 and -6 for my factorization:

Since I am multiplying to a "minus" six, I need factors that have opposite signs; that is, one factor will be "plus" and the other will be "minus". The coefficient on the middle term is a "plus" 1, so I know that the larger of the two factors (larger, that is, in terms of its absolute value) will get the "plus" sign. Since these opposite-signed numbers will be added together to get +1, then I need the two factors to be one unit apart.

The factor pairs for 6 are 1 and 6, and 2 and 3. The values in the second pair are one unit apart, so I know that I will be using 2 and 3.

Because I need to get a "plus" answer for the sum of the two factors, I'll need the larger of my two numbers to get the "plus" sign; namely, the 3 will be getting the "plus" sign (so the 2 gets the "minus" sign). Then my factorization is:

This looks just like the previous quadratic, except that now the middle term is "minus". The constant term is still "minus", so I still want factors with opposite signs. And the middle term's coefficient (other than its sign) is still 1, so I still want factors that are one unit apart. But this time the larger factor will get the "minus" sign.

All the other considerations remain the same. I still want factors of 6 that are one unit apart, so I'll still be using 2 and 3. But this time, the sign on the 3 will be a "minus":

In this quadratic, the constant term is "minus", so I still want factors of opposite signs.

However, this time, the middle term's coefficient (other than its sign) is 5 rather than 1, so now I want my two factors to be five units apart. And because the middle term's coefficient is "minus", the larger of my two factors will get the "minus" sign.

The factor pairs for six are 1 and 6, and 2 and 3. The numbers in the first pair of factors are five apart, so I'll be using the numbers 1 and 6. I'm adding to a "minus", so I'll give the larger of my factors the "minus" sign; that is, I'll be using +1 and -6 for this factorization:

There is one special case, by the way, for factoring quadratics. Back when you were factoring plain old numbers, there were some numbers that didn't factor, such as 5 or 13. Recall that these unfactorable numbers are called "prime" numbers. You could always multiply two whole numbers to get another whole number, but some whole numbers could not be divided to get two (non-trivial) whole numbers.

Similarly, there are quadratics that won't factor. You can multiply two binomials (without fractions) to get a quadratic (without fractions), but not all quadratics can be factored to get two (non-trivial) binomials. The terminology for such quadratics (or any un-factorable polynomial) is also "prime".

The leading coefficient is 1, so this is a case of simple factoring. Since the constant term is "minus", I'll be needing a "plus" factor and a "minus" such that, when I multiply them together, I get -6, but when I add them, I get +7. In other words, I'll need two factors that are seven units apart, with the larger factor getting the "plus" sign.

The factor pairs for 6 are 1 and 6, and 2 and 3. At first glance, I may assume that I should use 1 and 6, but -

One of the factors has to be "minus" in order to multiply to get a "minus" six! Trying the first factor pair of 1 and 6, the sum would be one of the following:

(-1) + 6 = 5

1 + (-6) = -5

Okay; neither way worked. So maybe I should be using the other factor pair...? Checking the addition with those numbers, I get:

(-2) + 3 = 1

2 + (-3) = -1

One might think, "Well, maybe I should should do the signs differently." But the only way to get factors of six to add up to seven is to give those factors the same sign. But if I do that, then the factors won't multiply to a "minus" six.

In other words, there is no pair of factors of -6 that will add to +7. And if something isn't factorable, then it's prime. So is a prime quadratic polynomial. In technical terms, it is "unfactorable over the integers", so-called because I couldn't find a pair of integers that would work. Then my answer is:

URL: https://www.purplemath.com/modules/factquad.htm

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A "quadratic" is a polynomial that looks like "", where "a", "b", and "c" are just numbers (and either of b and c, but never a, might be zero). For the easy case of factoring quadratic polynomials, we will need to find two numbers that will multiply to be equal the constant term c, and will also add up to equal b, the coefficient on the linear -term in the middle.

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Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

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https://www.youtube.com/watch?v=xJrdNSXQPRU

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😎Multiplying Polynomials Like a Boss! [fbt] (FAST)🤑

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My name is Karin Hutchinson. I am one of those CRAZY teachers who loves to teach Algebra.

After helping many middle school students (who despised math) learn to LOVE Algebra, I decided to try to help more students around the world get rid of their fears and frustrations of Algebra.

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Detailed Examples to Help You Understand Algebra Concepts

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My name is Karin Hutchinson. I am one of those CRAZY teachers who loves to teach Algebra. After helping many middle school students (who despised math) learn to LOVE Algebra, I decided to try to help more students around the world get rid of their fears and frustrations of Algebra.

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WHEN NUMBERS ARE ADDED OR SUBTRACTED, we call them terms. Like terms are terms that we could combine. Here, for example, is a sum of like terms:

5 x + 3 x − 2 x.

How many x 's are there? 5 + 3 − 2 equals 6 of them. We write:

5 x + 3 x − 2 x = 6 x.

We call 5, 3, and −2 the coefficients respectively of each term. Recall that in naming terms we include the minus sign.

Here, on the other hand, is a sum of unlike terms:

x² − 2 xy + y ²

There is no way that we could combine them.

The most elementary terms that we could combine are a and − a.

5 + (−5) = 0.

Adding like terms

In this sum-

2 x − 3 y + 4 x + 5 y

-the like terms are 2 x and 4 x, y and −5 y.

Upon combining, or adding, them:

234− 56− 2 x + y + x y = x y.

That has the effect of reducing the number of terms. Which is what we like.

We say that there are two terms "in" x and two "in" y. The preposition "in" indicates which are the like terms.

See Problem 12.

To add like terms, add their coefficients. The order in which the terms are written does not matter.

Problem 1 . 6 a − 3 b + c − d.

a) What number is the coefficient of a ?

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

6

b) What number is the coefficient of b ? −3

c) What number is the coefficient of 1. For c ? c = 1 ·c.

d) What number is the coefficient of −1. − d ? d = −1 ·d.

See Lesson 5.

It is the style in algebra not to write the coefficients 1 or −1.

Actually, the coefficient of any factor is all the remaining factors. Thus in the term 4 ab, the coefficient of a is 4 b; the coefficient of 4 a is b; and so on. In this term, x( x − 1), the coefficient of ( x − 1) is x.

Problem 2. In the expression 5 ayx, name the coefficient of

a) 5 5 5 x ay b) y ax c) yx a

d) 5 a xy e) 5 ayx

Problem 3. In this product 2( x + y) z

a) name each factor. 2, ( x + y), z

b) name the coefficient of 2( z. x + y)

Problem 4. What number is the coefficient of x?

b)

3x
4

3
4

3x
4

=

3
4

· x

Lesson 4.

Problem 5. How do we add like terms?

Add their coefficients.

Problem 6. Add like terms.

Again, we do not to write the coefficients 1 or −1.

k) 4= 2 x + y − 2 x + y x + 2 y

l) 3= −5 x − y − 8 x + 2 y x + y

m) 4 x² − 5 x² + x ² = 0

Problem 7. Add like terms.

b) 2 = 6 a + 3 b + 4 a − 5 ab a + 3 b − 5 ab.
Terms that you cannot combine, simply rewrite.

Problem 8. Remove parentheses and add like terms.

=

=

7 a − 9 b + 2 c

= a + 2 b + 4 c − 3 d −3 a + 8 b + 2 c − d

= −2 a + 10 b + 6 c − 4 d

= 4 x − 3 y + 3 y − 5 x + 5z − 4 x

= −5 x + 5z

d) (5 x y − 3 x + 2 y − 1) − (2 x y − 7 x − 8 y + 6)

= 5 xy − 3 x + 2 y − 1 −2 x y + 7 x + 8 y − 6

= 3 xy + 4 x + 10 y − 7

e) ( x − y) − ( y + xy − x) − (2 x − 4 x y − 2 y)

= x − y − y − x y + x − 2 x + 4 xy + 2 y

= 3 xy

f) (4 x² − 7 x − 3) − ( x² − 4 x + 1)

= 4 x² − 7 x − 3 − x² + 4 x − 1

= 3 x² − 3 x − 4

g) (6 x ³ + 4 x² − 2 x − 6) − (2 x ³ − 8 x² + x − 2)

= 6 x ³ + 4 x² − 2 x − 6 − 2 x ³ + 8 x² − x + 2

= 4 x ³ + 12 x² − 3 x − 4

Problem 9. 5 abc + 2 cba. Are those like terms?

Yes. The order of factors does not matter.
Upon combining them, we get 7 abc.
When writing the final answer, it is conventional to preserve the alphabetical order.

Problem 10. Add like terms.

a)

4= −5 xy − 9 yx xy

b)

8= 4 x − 5 x y − 4 x + 4 y x x − xy

c) 9 = 17 xyz + 3 yzx + 5 zxy xyz

d) 3 = −5 xy − 4 xyz + 3 x − 8 yx + 5 yzx − 9 x xy + xyz − 6 x

Problem 11. Add like terms.

c) 2 n + 4 − 2 n − 2 = 2

Problem 12. Add like terms, which are in ( x + 2). Do not remove parentheses.

c) ( x( x + 2) + 4( x + 2) = x + 4)( x + 2).

We added the coefficients.

d) ( x( x + 2) − ( x + 2) = x − 1)( x + 2).

Problem 13. Add like terms, which are in x or y. Add the coefficients.

a) ( px + qx = p + q) x.

b) ( ax + by − cx + dy = a − c) x + ( b + d) y.

c) (1 + ( x + ax = a) x. d) ax − a − 1) x. x =

e) (( a + b) x + cx = a + b + c) x.

f) (( a − b) x − cx = a − b − c) x.

f) ( a + b) x − ( b + a) x = 0.

Problem 14. Add like terms.

a) 34 a ² b ³ − 2 a b ² + a ³ b ² − 5 b ² a + b ³ a ² = a ² b ³ − 7 a b ² + a ³ b ².

b) 3 x y ² − xy + x ² y − y ² x + 2 y x ² + yx = x ² y.

*

In calculus, the student will not see any problem in the form "Subtract a from b." However, in certain standard exams that form tends to come up. Hence, the following rule.

The rule for subtraction

"Subtract a from b." Is that a − b or b − a ?

It is b − a. a is the number being subtracted. It is called the subtrahend. The subtrahend appears to the right of the minus sign -- before the word "from."

Example. Subtract 2 x − 3 from 5 x − 4

Solution. 2 x − 3 is the subtrahend.

(5 x − 4) − (2 x − 3)

=

5 x − 4 − 2 x + 3

=

3 x − 1.

Notice: The signs of the subtrahend change.

2 x − 3 changes to −2 x + 3.

We can therefore state the following rule for subtraction.

Change the signs of all the terms in the subtrahend.
Then add the like terms.

Problem 15. Subtract 4 a − 2 b from a + 3 b.

Change the signs of the subtrahend, and add:

a + 3 b − 4 a + 2 b = −3 a + 5 b.

Problem 16. Subtract x² − 5 x + 7 from 3 x² − 8 x − 2.

3 x² − 8 x − 2 − x² + 5 x − 7

=

2 x² − 3 x − 9.

Next Lesson: Linear equations
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Adding like terms - A complete course in algebra

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WHEN NUMBERS ARE ADDED OR SUBTRACTED, we call them terms. Like terms are terms that we could combine. Here, for example, is a sum of like terms: 5 x + 3 x − 2 x. How many x 's are there? 5 + 3 − 2 equals 6 of them.

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Solves your equations step-by-step and shows the work! This calculator will solve your problems.

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Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In this non-linear system, users are free to take whatever path through the material best serves their needs. These unique features make Virtual Nerd a viable alternative to private tutoring.

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This animated Math Shorts video from UEN explains the term scientific notation and provides several examples in converting extreme numbers to and from scientific notation. It also demonstrates how to multiply numbers using scientific notation. In the accompanying classroom activity, students practice writing numbers in scientific notation and develop real-world problems for each other to solve.

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Exponents' Rules

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ratios and proportions notes and exercises

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ratios and proportions notes and exercises

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For an explanation and everyday examples of using percentages generally see our page Percentages: An Introduction. For more general percentage calculations see our page Percentage Calculators.

To calculate the percentage increase:

First: work out the difference (increase) between the two numbers you are comparing.

Increase = New Number - Original Number

Then: divide the increase by the original number and multiply the answer by 100.

% increase = Increase ÷ Original Number × 100.

If your answer is a negative number, then this is a percentage decrease.

To calculate percentage decrease:

First: work out the difference (decrease) between the two numbers you are comparing.

Decrease = Original Number - New Number

Then: divide the decrease by the original number and multiply the answer by 100.

% Decrease = Decrease ÷ Original Number × 100
If your answer is a negative number, then this is a percentage increase.
If you wish to calculate the percentage increase or decrease of several numbers then we recommend using the first formula. Positive values indicate a percentage increase whereas negative values indicate percentage decrease.

Percentage Change Calculator

Use this calculator to work out the percentage change of two numbers

More: Percentage Calculators

Further Reading from Skills You Need

The Skills You Need Guide to Numeracy
This four-part guide takes you through the basics of numeracy from arithmetic to algebra, with stops in between at fractions, decimals, geometry and statistics.

Whether you want to brush up on your basics, or help your children with their learning, this is the book for you.

Examples - Percentage Increase and Decrease
In January Dylan worked a total of 35 hours, in February he worked 45.5 hours - by what percentage did Dylan's working hours increase in February?
To tackle this problem first we calculate the difference in hours between the new and old numbers. 45.5 - 35 hours = 10.5 hours. We can see that Dylan worked 10.5 hours more in February than he did in January - this is his increase. To work out the increase as a percentage it is now necessary to divide the increase by the original (January) number:

10.5 ÷ 35 = 0.3 (See our division page for instruction and examples of division.)

Finally, to get the percentage we multiply the answer by 100. This simply means moving the decimal place two columns to the right.
0.3 × 100 = 30 Dylan therefore worked 30% more hours in February than he did in January.
In March Dylan worked 35 hours again - the same as he did in January (or 100% of his January hours). What is the percentage difference between Dylan's February hours (45.5) and his March hours (35)?

First calculate the decrease in hours, that is: 45.5 - 35 = 10.5

Then divide the decrease by the original number (February hours) so:

10.5 ÷ 45.5 = 0.23 (to two decimal places).

Finally multiply 0.23 by 100 to give 23%. Dylan's hours were 23% lower in March than in February.

You may have thought that because there was a 30% increase between Dylan's January hours (35) and February (45.5) hours, that there would also be a 30% decrease between his February and March hours. As you can see, this assumption is incorrect.

The reason is because our original number is different in each case (35 in the first example and 45.5 in the second). This highlights how important it is to make sure you are calculating the percentage from the correct starting point.

Sometimes it is easier to show percentage decrease as a negative number - to do this follow the formula above to calculate percentage increase - your answer will be a negative number if there was a decrease. In Dylan's case the increase in hours between February and March is -10.5 (negative because it is a decrease). Therefore -10.5 ÷ 45.5 = -0.23. -0.23 × 100 = -23%.

Dylan's hours could be displayed in a data table as:

Calculating Values Based on Percentage Change

Sometimes it is useful to be able to calculate actual values based on the percentage increase or decrease. It is common to see examples of when this could be useful in the media.

You may see headlines like:

UK rainfall was 23% above average this summer.
Unemployment figures show a 2% decline.
Bankers' bonuses slashed by 45%.

These headlines give an idea of a trend - where something is increasing or decreasing, but often no actual data.

Without data, percentage change figures can be misleading.

Ceredigion, a county in West Wales, has a very low violent crime rate.

Police reports for Ceredigion in 2011 showed a 100% increase in violent crime. This is a startling number, especially for those living in or thinking about moving to Ceredigion.

However, when the underlying data is examined it shows that in 2010 one violent crime was reported in Ceredigion. So an increase of 100% in 2011 meant that two violent crimes were reported.

When faced with the actual figures, perception of the amount of violent crime in Ceredigion changes significantly.

In order to work out how much something has increased or decreased in real terms we need some actual data.

Take the example of " UK rainfall this summer was 23% above average " - we can tell immediately that the UK experienced almost a quarter (25%) more rainfall than average over the summer. However, without knowing either what the average rainfall is or how much rain fell over the period in question we cannot work out how much rain actually fell.

Calculating the actual rainfall for the period if the average rainfall is known.

If we know the average rainfall is 250mm, we can work out the rainfall for the period by calculating 250 + 23%.

First work out 1% of 250, 250 ÷ 100 = 2.5. Then multiply the answer by 23, because there was a 23% increase in rainfall.

2.5 × 23 = 57.5.
Total rainfall for the period in question was therefore 250 + 57.5 = 307.5mm.
Calculating the average rainfall if the actual amount is known.

If the news report states the new measurement and a percentage increase, " UK rainfall was 23% above average... 320mm of rain fell... ".

In this example we know the total rainfall was 320mm. We also know that this is 23% above the average. In other words, 320mm equates to 123% (or 1.23 times) of the average rainfall. To calculate the average we divide the total (320) by 1.23.

320 ÷ 1.23 = 260.1626. Rounded to one decimal place, the average rainfall is 260.2mm.

The difference between the average and the actual rainfall can now be calculated:
320 - 260.2 = 59.8mm.

We can conclude that 59.8mm is 23% of the average rainfall amount (260.2mm), and that in real terms, 59.8mm more rain fell than average.

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Percentage Change | Increase and Decrease

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Quickly learn to calculate the increase or decrease in percentage terms. Formula, real-life examples and percentage change calculator.

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15 minute math decimals

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tennessee adult education mathematics pre-ged

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Tennessee Adult Education Mathematics Pre-GED - ppt download

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What is percent? It is another way to write a part of a whole. It refers to the number of parts out of 100 equal parts. 6% means 6 parts out of 100 In other words: "Percent" literally means " per 100"

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Please browse through the entire lesson.