Last post, we talked about radical inequalities. In this post, we will talk about how to solve logarithmic inequalities. We’ll see logarithmic inequalities in forms such as or . In order to solve these inequalities, the goal will be to isolate the variable, just as in any inequality, and we will do this by getting rid of the log function. Let’s dive in and see how to solve logarithmic inequalities.

Steps to solve logarithmic inequalities:

1. Use algebraic manipulation to move anything that is not in the logarithmic expression to one side

2. Combine logarithmic expressions

3. Isolate the variable by getting rid of the logarithmic expression

Ex:

Ex:

4. Solve inequality

5. Get the range for the expression in the original log function

Ex:

Range:

6. Combine ranges

Let’s do an example step by step now.

First example (click
here):

Step 1: Use algebraic manipulation to move anything that is not in the logarithmic expression to one side

There’s nothing to move, so we can skip this step.

Step 2: Combine logarithmic expressions

Step 3: Isolate the variable by getting rid of the logarithmic expression

Step 4: Solve inequality

We can see that this is now a rational inequality. We won’t solve this step by step; I will show the answer after solving this inequality. If you are struggling with solving this inequality, visit the blog post on rational inequalities.

Step 5: Get the range for the expression in the original log function

Step 6: Combine ranges

, ,

That wasn’t too bad! Let’s see some more examples.

Second example (click
here):

Last example (click
here):

Solving logarithmic inequalities is not too difficult. Just remember to get the ranges inside the logarithmic expressions and to double check your work. For more help and practice on this topic visit Symbolab’s practice.

Until next time,

Leah

## Tuesday, November 29, 2016

## Tuesday, November 22, 2016

### High School Math Solutions – Inequalities Calculator, Radical Inequalities

Last post, we went over how to solve absolute value inequalities. For today’s post, we will talk about how to solve radical inequalities. Solving radical inequalities is easier than solving absolute value inequalities and require fewer steps. Let’s see the steps on how to solve these inequalities.

Steps:

Let’s see how to do one example step by step.

First example (click here):

\sqrt{5+x}-1<3

Step 1: Isolate the square root

\sqrt{5+x}<4

Step 2: Check that the inequality is true

Yes, this inequality is true the radical is not less than 0.

Step 3: Find the real region for the square root

5+x≥0

x≥-5

Step 4: Simplify and compute the inequality

\sqrt{5+x}<4

(\sqrt{5+x})^2<4^2

5+x<16

x<11

Step 5: Combine the ranges

x≥-5 and x<11

-5≤x<11

That wasn’t too difficult. Let’s see some more examples.

Second example (click here):

Last example (click here):

Solving radical inequalities isn’t too difficult, however, they require practice. For more practice examples check out Symbolab’s practice.

Until next time,

Leah

Steps:

- Isolate the square root
- Check that the inequality is true (i.e. not less than 0)
- Find the real region for the square root, (i.e. see when the expression inside square root is greater than or equal to 0)
- Simplify and compute the inequality
- Combine the ranges

Let’s see how to do one example step by step.

First example (click here):

\sqrt{5+x}-1<3

Step 1: Isolate the square root

\sqrt{5+x}<4

Step 2: Check that the inequality is true

Yes, this inequality is true the radical is not less than 0.

Step 3: Find the real region for the square root

5+x≥0

x≥-5

Step 4: Simplify and compute the inequality

\sqrt{5+x}<4

(\sqrt{5+x})^2<4^2

5+x<16

x<11

Step 5: Combine the ranges

x≥-5 and x<11

-5≤x<11

That wasn’t too difficult. Let’s see some more examples.

Second example (click here):

Last example (click here):

Solving radical inequalities isn’t too difficult, however, they require practice. For more practice examples check out Symbolab’s practice.

Until next time,

Leah

## Thursday, November 3, 2016

### High School Math Solutions – Polynomials Calculator, Dividing Polynomials (Long Division)

Last post, we talked dividing polynomials using factoring and splitting up the fraction. In this post, we will talk about another method for dividing polynomials, long division. Long division with polynomials is similar to the basic numerical long division, except we are dividing variables. This is where it gets tricky. I will talk about the steps to dividing polynomials using long division to help make the process easier and go into detail.

3. Write 0 as the coefficient for missing terms in the dividend

Steps for polynomial long division:

1. Organize each polynomial by higher order

- We want to make sure that each polynomial is written in order of the variable with the highest exponent to the variable with the lowest exponent

- you can skip this step if they are already in high order

2. Set up in long division form

- The denominator becomes the divisor and the numerator becomes the dividend

3. Write 0 as the coefficient for missing terms in the dividend

- Since we’ve put in order the terms based on their exponent, we can see which terms are missing (i.e. x^4+x^2 we can see we are missing an x^3 term so we will add that in to its proper spot and make the coefficient 0)

- This will help you with step 6, so you don’t subtract the wrong terms

4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)

- You can skip this step if there are no missing terms

5. Multiply the divisor by that term

- This is allows us to see what we need to multiply the divisor by to get rid of the first term of the dividend

6. Subtract this from the dividend

- Write the term down on top of line where the term that is getting eliminated is

7. Repeat steps 4-6 until you get a remainder

- This gives you a new polynomial to work with

8. Put the remainder over the divisor to create a fraction and add it to the new polynomial

- When you repeat step 4, move onto the newest first term from step 6

This may seem a bit confusing, so we will go through two examples step by step to understand better how to solve these problems.

First example (click here):

\frac{(x^4+6x^2+2)}{(x^2+5)}

1. Organize each polynomial by high order

We can skip this step because the polynomials are already in high order

2. Set up in long division form

3. Write 0 as the coefficient for missing terms in the dividend

4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)

\frac{x^4}{x^2}= x^2

5. Multiply the divisor by that term

x^2∙(x^2+5)=x^4+5x^2

6. Subtract this from the dividend

7. Repeat step 4-6 until you get a remainder

\frac{x^2}{x^2} =1

8. Put the remainder over the divisor to create a fraction and add it to the new polynomial

x^2+1+\frac{(-3)}{(x^2+5)}

Last example (click here):

\frac{(2x^2-18+5x)}{(x+4)}

1. Organize each polynomial by high order

\frac{(2x^2+5x-18)}{(x+4)}

2. Set up in long division form

3. Write 0 as the coefficient for missing terms in the dividend

We can skip this step because there are no missing terms.

4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)

\frac{2x^2}{x}=2x

5. Multiply the divisor by that term

2x(x+4)=2x^2+8x

6. Subtract this from the dividend

7. Repeat steps 4-6 until you get the remainder

\frac{(-3x)}{x}=-3

8. Put the remainder over the divisor to create a fraction and add it to the new polynomial

2x-3+\frac{(-6)}{(x+4)}

Dividing polynomials using long division is very tricky. It is so easy to skip an exponent, have an algebraic error, and forget a step. This is why practicing this type of problem is so important. The only way to get better at it is to keep practicing it. Check out Symbolab’s Practice for practice problems and quizzes.

Until next time,

Leah.

First example (click here):

\frac{(x^4+6x^2+2)}{(x^2+5)}

1. Organize each polynomial by high order

We can skip this step because the polynomials are already in high order

2. Set up in long division form

3. Write 0 as the coefficient for missing terms in the dividend

4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)

\frac{x^4}{x^2}= x^2

5. Multiply the divisor by that term

x^2∙(x^2+5)=x^4+5x^2

6. Subtract this from the dividend

7. Repeat step 4-6 until you get a remainder

\frac{x^2}{x^2} =1

8. Put the remainder over the divisor to create a fraction and add it to the new polynomial

x^2+1+\frac{(-3)}{(x^2+5)}

Last example (click here):

\frac{(2x^2-18+5x)}{(x+4)}

1. Organize each polynomial by high order

\frac{(2x^2+5x-18)}{(x+4)}

2. Set up in long division form

3. Write 0 as the coefficient for missing terms in the dividend

We can skip this step because there are no missing terms.

4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)

\frac{2x^2}{x}=2x

5. Multiply the divisor by that term

2x(x+4)=2x^2+8x

6. Subtract this from the dividend

7. Repeat steps 4-6 until you get the remainder

\frac{(-3x)}{x}=-3

8. Put the remainder over the divisor to create a fraction and add it to the new polynomial

2x-3+\frac{(-6)}{(x+4)}

Dividing polynomials using long division is very tricky. It is so easy to skip an exponent, have an algebraic error, and forget a step. This is why practicing this type of problem is so important. The only way to get better at it is to keep practicing it. Check out Symbolab’s Practice for practice problems and quizzes.

Until next time,

Leah.

## Wednesday, October 26, 2016

### High School Math Solutions – Polynomials Calculator, Dividing Polynomials

In the last post, we talked about how to multiply polynomials. In this post, we will talk about to divide polynomials. When we divide polynomials, we can write these problems in the form of a fraction. This allows us to reduce the fraction and get our answer. Although this method doesn’t work every time we divide polynomials, when it does, it is a quick, simple method.

Once written in the form of a fraction, there are two ways to reduce the fraction. One method is to split the fraction up, so there is one term per fraction, having the denominator be the same. The other method is to factor the numerator and denominator and cancel out terms that are the same in the numerator and denominator.

Let’s see some examples to better understand how to solve these problems.

We will use the splitting method first for our first example.

First example (click here):

Next example:

We will use our same problem above, but use factoring to solve the problem.

4. Simplify

Last example (click here):

4. Simplify

As you can see, factoring is a big part of dividing polynomials. Before attempting these problems, make sure you’ve mastered factoring. Dividing polynomials by these two methods is pretty simple. However, these methods may not work for certain division problems. Next blog post, we will talk about another method to use, when these methods don’t work. For more practice and help on this topic, checkout Symbolab’s Practice.

Until next time,

Leah

Once written in the form of a fraction, there are two ways to reduce the fraction. One method is to split the fraction up, so there is one term per fraction, having the denominator be the same. The other method is to factor the numerator and denominator and cancel out terms that are the same in the numerator and denominator.

Let’s see some examples to better understand how to solve these problems.

We will use the splitting method first for our first example.

First example (click here):

(32x^4-56x^2)\div 8x1. Rewrite the problem in the form of a fraction

\frac{32x^4-56x^2}{8x}2. Split the fraction

\frac{32x^4}{8x}-\frac{56x^2}{8x}3. Reduce the fractions

\frac{8\cdot 4x^4}{8x}-\frac{8\cdot7x^2}{8x}

4x^3-7x

Next example:

We will use our same problem above, but use factoring to solve the problem.

(32x^4-56x^2)\div 8x1. Rewrite the problem in the form of a fraction

\frac{32x^4-56x^2}{8x}2. Factor the numerator and denominator

\frac{8x^2(4x^2-7)}{8x}3. Cancel out common factors

4. Simplify

x(4x^2-7)

4x^3-7x

Last example (click here):

(2x^3+11x^2+5x)\div(2x^2+x)1. Rewrite the problem in the form of a fraction

\frac{2x^3+11x^2+5x}{2x^2+x}2. Factor the numerator and denominator

\frac{(2x+1)(x+5)x}{(2x+1)x}3. Cancel out common factors

4. Simplify

x+5

As you can see, factoring is a big part of dividing polynomials. Before attempting these problems, make sure you’ve mastered factoring. Dividing polynomials by these two methods is pretty simple. However, these methods may not work for certain division problems. Next blog post, we will talk about another method to use, when these methods don’t work. For more practice and help on this topic, checkout Symbolab’s Practice.

Until next time,

Leah

### Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics

Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process of finding factors of a number or expression, where we find what multiplies together to make the expression or number. In today’s blog post, we will talk about how to factor simple expressions and quadratics.

Given a simple expression, ax+b, pull out the greatest common factor from the expression. Pretty simple!

Here’s an example (click here):

Quadratics have the form: (ax)^2+bx+c, where a, b, and c are numbers.

Here are the steps for factoring quadratics:

Here’s an example, where 1 is the leading coefficient (click here):

Here’s an example when 1 isn’t the leading coefficient (click here):

The best advice for factoring quadratics is to practice factoring as many quadratics as you can. The more you practice factoring, the faster you will get. Soon, you’ll be able to skip the steps and factor it all in your head within seconds. For more help or practice on the topic, check out Symbolab’s Practice.

Until next time,

Leah

__Factoring simple expressions –__Given a simple expression, ax+b, pull out the greatest common factor from the expression. Pretty simple!

Here’s an example (click here):

Factor\:2x+6

2x+2\cdot3

2(x+3)

__Factoring quadratics –__Quadratics have the form: (ax)^2+bx+c, where a, b, and c are numbers.

Here are the steps for factoring quadratics:

- Find u and v such that u∙v=a∙c and u+v=b

This means u and v are factors of a∙c that when added together equal b - Rewrite the expression as (ax^2+ux)+(vx+c)
- Factor out what you can from each parentheses
- Factor out a common term
- Check by multiplying the factors together (FOIL)

Here’s an example, where 1 is the leading coefficient (click here):

Factor\:x^2-5x+61. Find u and v such that u∙v=a∙c and u+v=b

a∙c=1∙6=6

6 can be written as the product of 1 and 6, -1 and -6, 3 and 2, or of -3 and -2. We need to pick the factors of 6 that equal -5.

-3+(-2)=-52. Rewrite the expression as (ax^2+ux)+(vx+c)

-3∙-2=6

u=-3,\:v=-2

(x^2+(-3x))+(-2x+6)3. Factor out what you can from each parentheses

(x^2-3x)+(-2x+6)

(x^2-3x)+(-2x+6)

x(x-3)+2(-x+3)

x(x-3)-2(x-3)

We factored out a -1 from (-x+3) on the last step because we want the expressions inside the parentheses to be the same for step 4.4. Factor out a common term

x(x-3)-2(x-3)5 .Check by multiplying the factors together

(x-3)(x-2)

(x-3)(x-2)=x^2-2x-3x+6=x^2-5x+6

Here’s an example when 1 isn’t the leading coefficient (click here):

Factor\:2x^2+x-61. Find u and v such that u∙v=a∙c and u+v=b

a∙c=2∙-6=-12

Factors of -12: -12 and 1, 12 and -1, -6 and 2, 6 and -2, -4 and 3, 4 and -32. Rewrite the expression as (ax^2+ux)+(vx+c)

4-3=1

4∙-3=-12

u=-4,\:v=3

(2x^2-4x)+(3x-6)3. Factor out what you can from each parentheses

(2x^2-4x)+(3x-6)4. Factor out a common term

2x(x-2)+3(x-2)

(x-2)(2x+3)5. Check by multiplying the factors together

(x-2)(2x+3)=2x^2+3x-4x-6=(2x)^2-x-6

The best advice for factoring quadratics is to practice factoring as many quadratics as you can. The more you practice factoring, the faster you will get. Soon, you’ll be able to skip the steps and factor it all in your head within seconds. For more help or practice on the topic, check out Symbolab’s Practice.

Until next time,

Leah

### Symbolab Study Groups…groups that work

stud•y group

plural noun:

a group of people who meet to study a particular subject and then report their findings or recommendations.

Study groups are great; more brainpower, boost motivation, support system. But let’s be honest, it’s not always the most effective way to learn. Sessions can turn into social events, schedule doesn’t work for everyone, we’re not always prepared for the sessions…

Symbolab Groups is a game changer. There’s no better way to connect, share notes, and work through difficult problems together. Symbolab Groups is a stress free study environment. You get all the benefits of a study group minus the distractions.

Symbolab helps you stay focused. No need to worry about missing out on group meetings, or not taking notes. Stuck on a problem? Just ask, your friends can help you instantly. Share problems, exercises and graphs, start a discussion, answer questions.

Symbolab Groups are there for you, always.

Cheers,

Michal

*noun*plural noun:

**study groups**a group of people who meet to study a particular subject and then report their findings or recommendations.

Study groups are great; more brainpower, boost motivation, support system. But let’s be honest, it’s not always the most effective way to learn. Sessions can turn into social events, schedule doesn’t work for everyone, we’re not always prepared for the sessions…

Symbolab Groups is a game changer. There’s no better way to connect, share notes, and work through difficult problems together. Symbolab Groups is a stress free study environment. You get all the benefits of a study group minus the distractions.

Symbolab helps you stay focused. No need to worry about missing out on group meetings, or not taking notes. Stuck on a problem? Just ask, your friends can help you instantly. Share problems, exercises and graphs, start a discussion, answer questions.

Symbolab Groups are there for you, always.

Cheers,

Michal

## 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.

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:

When multiplying polynomials, you may come across multiplying variables with exponents by variables with exponents. In this case, we use this exponent rule:

Let’s see some examples to understand how to multiply polynomials.

First example (click here):

1. Use FOIL identity

4. Add like terms

Next example (click here):

1. Use (a+b)(a-b)=a^2-b^2

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

__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^2Multiplying 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)2. Multiply terms

2x\cdot 5x+2x\cdot -6+(-1)\cdot 5x+(-1)\cdot -6

10x^2-12x-5x+63. 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)2. Simplify

(2x^2 )^2-6^2

4x^4-6^2Last example (click here):

4x^4-36

(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 -13. Group like terms

2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6

2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-64. Add like terms

2x^4-3x^3+4x^3+6x^2-6x^2-2x^2+12x+3x-6

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

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