Wednesday, September 28, 2016

Middle School Math Solutions – Polynomials Calculator, Multiplying Polynomials

Multiplying polynomials can be tricky because you have to pay attention to every term, not to mention it can be very messy. There are a few ways of multiplying polynomials, depending on how many terms are in each polynomial. In this post, we will focus on how to multiply two term polynomials and how to multiply two or more term polynomials.

Multiply two term polynomials

When multiplying polynomials with two terms, you use the FOIL method. The FOIL method only works for multiplying two term polynomials. FOIL stands for first, outer, inner, last. This lets you know the order of how to distribute and multiply the terms. Let’s see how it works.


After FOILing, multiply the terms, group like terms, and add like terms if there are any.
Here is another helpful identity to use when multiplying two term polynomials:
(a+b)(a-b)=a^2-b^2
Multiplying these polynomials is pretty simple because if you memorize these identities then you just plug in the values and have an answer.

Multiplying multiple term polynomials

You cannot use the FOIL method to multiply these polynomials. Instead, you have to multiply each term in one polynomial by each term in the other. You can do this by multiplying each term of one polynomial by the other polynomial. This can be tricky because it is easy to miss one term. When we do examples of this, it will become easier to understand how to solve them.

When multiplying polynomials, you may come across multiplying variables with exponents by variables with exponents. In this case, we use this exponent rule:
x^n\cdot x^m=x^(n+m)
For this rule, the base or variable must be the same. When multiplying variables with exponents, you add the exponents together.


Let’s see some examples to understand how to multiply polynomials.
First example (click here):
(2x-1)(5x-6)
We will use the FOIL method to solve this.
1.   Use FOIL identity
(2x-1)(5x-6)
2x\cdot 5x+2x\cdot -6+(-1)\cdot 5x+(-1)\cdot -6
2.   Multiply terms
10x^2-12x-5x+6
3.   Group like terms
10x^2-12x-5x+6
 (Luckily, everything was already grouped together)

4.   Add like terms
10x^2-17x+6

Next example (click here):
(2x^2+6)(2x^2-6)
Here, we can use another one of the identities for multiplying two term polynomials.
1.   Use (a+b)(a-b)=a^2-b^2
(2x^2+6)(2x^2-6)
(2x^2 )^2-6^2
2.   Simplify
4x^4-6^2
4x^4-36
Last example (click here):
(x^2+2x-1)(2x^2-3x+6)
1.   Multiply each term in one polynomial by the other polynomial
x^2 (2x^2-3x+6)+2x(2x^2-3x+6)-1(2x^2-3x+6)
2.   Distribute and multiply
2x^2\cdot x^2-3x\cdot x^2+6\cdot x^2+2x^2\cdot 2x-3x\cdot 2x+6\cdot 2x+2x^2\cdot -1-3x\cdot -1+6\cdot -1
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
3.   Group like terms
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
2x^4-3x^3+4x^3+6x^2-6x^2-2x^2+12x+3x-6
4.   Add like terms
2x^4+x^3+6x^2-6x^2-2x^2+12x+3x-6
2x^4+x^3-2x^2+12x+3x-6
2x^4+x^3-2x^2+15x-6

Multiplying polynomials looks intimidating, but as long as you keep your work neat and double check your work, it should be pretty easy. Practice will be one of the biggest things that will help you. The more you practice, the easier multiplying polynomials will be because you will get the hang and flow of how to multiply them. Check out Symbolab’s Practice for more help and practice.

Until next time,

Leah

Tuesday, September 20, 2016

Middle School Math Solutions – Polynomials Calculator, Subtracting Polynomials

In the previous post, we talked about how to add polynomials. In this post, we will talk about subtracting polynomials. The key to subtracting polynomials is to make sure that you distribute the minus sign to the expression in the parentheses. Imagine that instead of – there is a -1. This should help you visualize why you are distributing the minus sign. Once the minus sign is distributed, the minus sign will turn into a plus sign. Then, you’ll be able to add the polynomials together. Let’s see the steps for subtracting polynomials.

Steps: 
  1. Distribute the negative sign
    • Include this step only if there is a minus sign in front of a polynomial in parentheses
  2. Remove the parentheses
    • Include this step if there are polynomials in parentheses
  3. Group like terms
    • Put together and order like terms (terms with the same variables and the same exponent)
  4. Add like terms 
    • Add the coefficients of the like terms

Let’s see some examples to better understand distributing the minus sign.



First example (click here):
(x^2+2x-1)-(2x^2-3x+6)
1.  Distribute the negative sign
(x^2+2x-1)-(2x^2-3x+6)
(x^2+2x-1)-1(2x^2-3x+6)
(x^2+2x-1)+(-2x^2+3x-6)
When we imagine that there is a -1 instead of – outside the parentheses, it is easy to see and remind ourselves that we have to distribute the negative. Distributing the negative sign allows that minus sign to turn into a plus sign.

2.   Remove the parentheses
x^2+2x-1+-2x^2+3x-6
x^2+2x-1-2x^2+3x-6
3.  Group like terms
x^2+2x-1-2x^2+3x-6
x^2-2x^2+2x+3x-1-6
4.  Add like terms
x^2-2x^2+2x+3x-1-6
-x^2+5x-7


Second example (click here):
(2x^3+2x-1)-(2x^2-5x-6)
1.  Distribute the negative sign
(2x^3+2x-1)-(2x^2-5x-6)
(2x^3+2x-1)-1(2x^2-5x-6)
(2x^3+2x-1)+(-2x^2+5x+6)
2.  Remove the parentheses
2x^3+2x-1+-2x^2+5x+6
2x^3+2x-1-2x^2+5x+6
3.  Group like terms
2x^3+2x-1-2x^2+5x+6
2x^3-2x^2+2x+5x-1+6
4.  Add like terms
2x^3-2x^2+7x-1+6
2x^3-2x^2+7x+5



Last example (click here):
(4x^3-x^2+x-2)-(-x^2+3)
1.  Distribute the negative sign
(4x^3-x^2+x-2)-(-x^2+3)
(4x^3-x^2+x-2)-1(-x^2+3)
(4x^3-x^2+x-2)+(x^2-3)
2.  Remove the parentheses
4x^3-x^2+x-2+x^2-3
3.  Group like terms
4x^3-x^2+x-2+x^2-3
4x^3-x^2+x^2+x-2-3
4.  Add like terms
4x^3+x-2-3
4x^3+x-5
Subtracting polynomials takes some practice to get the hang of distributing and remembering to distribute. Once you’ve mastered adding polynomials, subtracting them should be simple. For more practice, check out Symbolab’s Practice.

Until next time,

Leah




Monday, August 15, 2016

Middle School Math Solutions – Polynomials Calculator, Adding Polynomials

A polynomial is an expression of two or more algebraic terms, often having different exponents. Adding polynomials is pretty simple. The only tricky part is that there are different terms. The key is to add the like terms. Like terms are terms that have the same variable with the same exponent, even if they have different coefficients. There are three steps to adding polynomials:
  1. Remove parentheses - only include this step if the polynomials are in parentheses 
  2. Group like terms - put together and order like terms (terms with the same variable and same exponent)
  3. Add like terms - add the coefficients of the like terms

Let’s check out some examples step by step.

First example (click here):
(x^2+2x-1)+(2x^2-3x+6)
1. Remove parentheses
x^2+2x-1+2x^2-3x+6
2. Group like terms
x^2+2x^2+2x-3x-1+6
When I group like terms, I like to group and put the terms in order from the largest exponent to the smallest exponent.

3. Add like terms
3x^2+2x-3x-1+6
3x^2-x-1+6
3x^2-x+5

Second example (click here):
(2x^3+4x+7)+(2x^2+5x-6)
1. Remove parentheses
2x^3+4x+7+2x^2+5x-6
2. Group like terms
2x^3+2x^2+4x+5x+7-6
3. Add like terms
2x^3+2x^2+9x+1

Last example (click here):
4x+(4x-2)+(x^2-3)
1. Remove parentheses
4x+4x-2+x^2-3
2. Group like terms
x^2+4x+4x-2-3
3. Add like terms
x^2+8x-2-3
x^2+8x-5

Adding polynomials is pretty simple and requires just a few steps. The trickiest part about adding polynomials is making sure that you pay attention the exponents. It is easy to overlook and confuse a 2 for a 3. For more practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Thursday, June 9, 2016

High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part II

Last post we talked about absolute value inequalities with one absolute value expression. In this post, we will learn how to solve absolute value inequalities with two absolute value expressions. To do this, we will use some concepts from a previous post on rational functions. This can get a little tricky and confusing, so take time to read everything carefully.

I’ll show you the steps to solving these inequalities and then I will go through one example step by step.

Steps:
  1. For each absolute value expression, figure out their negative and positive ranges.
  2. Combine ranges if needed
  3. For each range, evaluate the absolute value inequality
  4. Combine the ranges

Here’s our first example (click here):
\frac{|3x+2|}{|x-1|} >2
Step 1: For each absolute value expression, figure out their negative and positive ranges
3x+2 x-1
Positive (≥0) 3x+2≥0
Range:x\ge-\frac{2}{3}
Positive absolute value expression:|3x+2|=3x+2
x-1≥0
Range:x≥1
Positive absolute value expression:|x-1|=x-1
Negative (<0) 3x+2<0
Range:x\le-\frac{2}{3}
Positive absolute value expression:|3x+2|=-(3x+2)
x-1<0
Range:x<1
Positive absolute value expression:|x-1|=-(x-1)


Step 2: Combine ranges if needed
Let’s see our ranges: x≥-\frac{2}{3},\quad x≥1,\quad x<-\frac{2}{3},\quad x<1
Now let’s combine any ranges if possible: x<-\frac{2}{3},\quad -\frac{2}{3}≤x<1, \quad x≥1

Step 3: For each range, evaluate the absolute value inequality

For x<-\frac{2}{3}:

\frac{-(3x+2)}{-(x-1)}>2 We use the negative absolute value expressions because for both expressions the range is less than their negative range.
\frac{3x+2}{x-1}>2
\frac{x+4}{x-1}>0 Now, we see that this is just a rational inequality. So we solve the rational inequality
x<-4\:or\:x>1 The steps aren’t shown to get the answer. You can refer back to the rational inequalities blog if you forget how to solve it

For -\frac{2}{3}≤x<1:

\frac{3x+2}{-(x-1)}>2 We use the negative absolute value expression for the |x-1| because the range is less than its range.
\frac{5x}{1-x}>0
0<x<1

For x≥1:
\frac{3x+2}{x-1}>2
\frac{x+4}{x-1}>2
x<-4\:or\:x>1

Step 4: Combine the ranges

(x<-\frac{2}{3}\:and\:x<-4\:or\:x>1)\:or\:(-\frac{2}{3}≤x<1\:and\:0<x<1)\:or\:(x≥1\:and\:x<-4\:or\:x>1)

Answer: x<-4 or x>1 or 0<x<1


Hopefully, that wasn’t too complicated. Now let’s see one more example.

Last example (click here):




As you can see, this topic can be very confusing and tricky. The more you practice, the more natural it will become. For more practice examples check out Symbolab’s practice.

Until next time,
Leah


High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part I

Last post, we learned how to solve rational inequalities. In this post, we will learn how to solve absolute value inequalities. There are two ways to solve these types of inequalities depending on what the inequality sign is. Let’s see how to solve them . . .

Case I (< or ≤):
Given |x|<a, we read this as “x is less than a units for 0.” On a number that looks like:


The red line is the solution for x. Remember because it is “less than,” a is not included in the points (this is why the circle is not filled in).

So we rewrite

The solution is in this form: -a<x<a.
The same goes for ≤. Just substitute ≤ for < and fill in the circles in the number line.

Case II(> or ≥):
Given |x|>a, we read this as “ x is more than a units from 0.” On a number line, that looks like:
The red line is the solution for x. Remember because it is “greater than.” A is not included in the points.

So we rewrite

The solution is 2 inequalities, NOT one. It is in the form: x<-a or x>a.
The same goes for ≥. Just substitute ≥ for > and fill in the circles in the number line.


NOTE: Remember to carry out any algebraic manipulations outside of the absolute value before continuing, for example:
2|x|-2>4
|x|>3
ANOTHER NOTE: Make sure a is positive or else there is no solution.


Let’s see some examples . . .


First example (click here):


Notice that the algebraic manipulations to be done are inside the absolute value, so that is done last.


Last example (click here):


That wasn’t too complicated. Next post, we will learn how to solve absolute value inequalities with two absolute value expressions in it. Make sure you practice because next post the problems will become trickier.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Rational Inequalities

Last post, we talked about solving quadratic inequalities. In this post, we will talk about rational inequalities. Let’s recall rational functions are an algebraic fraction that contain polynomials in the numerator and denominator. Solving rational inequalities is a little different than solving quadratic inequalities. Both share the concept of a table and testing values.

Let’s see how to solve rational inequalities.

Here the steps:
  1. Move everything to one side of the inequality sign
  2. Simplify the rational function 
  3. Find the zeros from the numerator and undefined points in the denominator
  4. Derive intervals
  5. Find the sign of the rational function on each interval
  6. Select the proper inequality

Let’s go through our first example step by step to understand the concept better.


Here’s our first example (click here):
\frac{x}{x-3}<4
Step 1: Move everything to one side of the inequality sign
\frac{x}{x-3}<4
\frac{x}{x-3}-4<0
\frac{-3x+12}{x-3}<0
Make sure you combine everything into one rational function.

Step 2: Simplify the rational function
\frac{-3x+12}{x-3}<0
\frac{3(-x+4)}{x-3}<0
It’s already simplified. Nothing to cancel out.

Step 3: Find the zeros from the numerator and undefined points in the denominator
-3x+12=0                                                        x-3=0
x=4                                                                     x=3
          Zero                                                                     Undefined point

Step 4: Derive intervals

Step 5: Find the sign of the rational function on each interval

Table Header x\lt3 3\ltx\lt4 x\gt4
\frac{-3x+12}{x-3} \frac{-3(0)+12}{(0)-3}=-4 \frac{-3(3.5)+12}{(3.5)-3}=3 \frac{-3(5)+12}{(5)-3}=-\frac{3}{2}
Sign \quad\quad\quad\quad- \quad\quad\quad\quad+ \quad\quad\quad\quad-

We changed the format of the intervals to inequalities. We pick a number in the interval, plug the number in the rational function and see what sign the answer is (negative or positive).

Step 6: Select the proper inequality
\frac{-3x+12}{x-3}<0
x<3\quad\:and\quad\:x>4
We refer back to the original inequality and see which inequality satisfies the original inequality. We are looking for a number that produces a negative number. We refer back to the table and see that x<3 and x>4 satisfy this.

Alright, that was a mouthful. Let’s see some more examples now…


Second example (click here):


Last example (click here):



Make sure you double check your work for calculation errors because that is where it’s very easy to make a mistake. For more practice, check out Symbolab’s practice.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Quadratic Inequalities

We’ve learned how to solve linear inequalities. Now, it’s time to learn how to solve quadratic inequalities. Solving quadratic inequalities is a little harder than solving linear inequalities. Let’s see how to solve them.

There are a couple ways to solve quadratic inequalities depending on the inequality. I’ll focus on explaining the more complicated version.

We’re given the quadratic inequality: x^2+2x-8\le0

Here are the steps to solving it:
  1. Move everything to one side of the inequality sign
  2. Set the inequality sign to an equal sign and solve for x
  3. Create three intervals
  4. Pick a number in each inequality and see if it satisfies the original inequality
  5. Select the proper inequality

Now we will go through this example step by step to understand a little better how to solve it.

Step 1: Make sure you start with 0 on one side
x^2+2x-8\le0
Looks good!

Step 2: Set the inequality sign to an equal sign and solve for x
x^2+2x-8=0
(x-2)(x+4)=0
x=2\:and\:x=-4
Step 3: Create three intervals


We are able to pick three intervals from looking at the number and seeing where the function crosses the x-axis (i.e. where the function is equal to 0).


Step 4: Pick a number in each inequality and see if it satisfies the original inequality
x<-4               -4<x<2               x>2
Table Header x<-4 -4<x<2 x>2
x^2+2x-8 (-5)^2+2(-5)-8=7 (0)^2+2(0)-8=-8 (3)^2+2(3)-8=7
Sign \quad\quad\quad\quad+ \quad\quad\quad\quad- \quad\quad\quad\quad+

We’ve turned the intervals into inequalities. Then, we picked a number in each inequality to see if it satisfied the original inequality, x^2+2x-8\le0.

Step 5: Select the proper inequality
-4<x<2
-4\lex\le2
We’ve selected -4<x<2 because it satisfies the original inequality because the quadratic is negative when x is between -4 and 2. However, don’t forget that the original equality contains the ≤ symbol so that means x can equal 0 too. We change the inequality signs because we know -4 and 2 are the zeros of the quadratic.


Let’s see some more examples…

Second example (click here):


Last example… This one is simple (click here):


Hopefully, that wasn’t too hard! Solving quadratic inequalities sometimes require patience to write everything out. For more help check out Symbolab’s practice.

Until next time,

Leah