Wednesday, November 15, 2017

Middle School Math Solutions – Expand Calculator, Perfect Squares

The perfect square formula is an application of the FOIL method that will help you calculate the square of a binomial quickly.

Let’s take a look at the formula:

                                                     (a+b)^2=a^2+2ab+b^2

Here’s the proof:

                                                    (a+b)^2=(a+b)(a+b)
                                                                   =a^2+ab+ab+b^2
                                                                   =a^2+2ab+b^2

Now, let’s see some examples using the perfect squares formula.

First example (click here):

                                                         Expand (x+1)^2

1. Apply the formula

                                                              a=xb=1

                                                 (x+1)^2=x^2+2∙x∙1+1^2
                                                                =x^2+2x+1

Second example (click here):

                                                         Expand (5-x)^2

1. Apply the formula

                                                             a=5b=-x

                                             (5-x)^2=5^2-2∙5∙(-x)+(-x)^2
                                                            =25+10x+x^2

Last example (click here):

                                                      Expand (s^2+4p)^2

1. Apply the formula

                                                            a=s^2, b=4p

                                        (s^2+4p)^2=(s^2)^2+2∙s^2∙4p+(4p)^2
                                                           =s^4+8s^2 p+16p^2

The more you practice these problems, the faster you will be able to do them. I’ve used this formula so much, that now, I don’t need it since I can do it all mentally. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Tuesday, November 7, 2017

Middle School Math Solutions – Expand Calculator, Two Squares

The difference of two squares is an application of the FOIL method (refer to our blog post on the FOIL method).   The difference of two squares is a number or term squared subtracted from another number or term squared. We get it when we multiply two binomials, where the terms in the binomials are the same, except one of the terms is subtracted instead of being added.

Let’s see this in a formula:
                                                             (a+b)(a-b)=a^2-b^2

Here’s the proof using the FOIL method:
                                                      (a+b)(a-b)=a^2-ab+ab-b^2
                                                                                =a^2-b^2

Not too complicated, so let’s see some examples.

First example (click here):
                                                             Expand (x+2)(x-2)

1. Apply the formula
                                                                        a=x,b=2

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

This next one is a little more complicated.

Next example (click here):
                                                           Expand (-y+2x)(y+2x)

1. Rewrite the problem
                                                      (-y+2x)(y+2x)=(2x-y)(2x+y)
                                                                                      =(2x+y)(2x-y)

2. Apply the formula

                                                                    a=2x,b=y

                                                     (2x+y)(2x-y)=(2x)^2-y^2
                                                                                   =4x^2-y^2

For our last example, we will see an application of the difference of two squares formula.

Last example (click here):
                                                                        23∙17

1. Rewrite the numbers
                                                                (20+3)(20-3)

2. Apply the formula
                                                                     a=20,b=3

                                                          (20+3)(20-3)=20^2-3^2
                                                                                        =400-9
                                                                                        =391

As you can see, this formula is simple, but very helpful. If you need more help or practice with this formula, check out Symbolab’s Practice.

Until next time,

Leah

Tuesday, October 31, 2017

Middle School Math Solutions – Expand Calculator, Polynomial Multiplication

As of right now, you should be familiar with the distributive law and the FOIL method. We will now use both of them and apply them to our next topic, polynomial multiplication.

Recall: A polynomial is an expression of many (poly-) terms (-nomial) that are added and/or subtracted together. The terms include coefficients, variables, and POSITIVE exponents.

Examples: 4x^2 ,2x^3+5x^5 ,\frac{2}{3}x-1

There are a variety of polynomial multiplication problems. They might look intimidating at first, but we will go over how to solve them step by step using the distributive law and FOIL method, so you can see how to do them easily.

First example (click here):

                                                             4x(3x^2+2x-1)

1. Distribute the 4x

                                          4x(3x^2+2x-1)=4x∙3x^2+4x∙2x+4x∙(-1)
We applied the distributive law here for a polynomial with three terms                  (a(b+c+d)=ab+ac+ad)
2. Simplify

                                                4x(3x^2+2x-1)=12x^3+8x^2-4x
We multiplied the terms.
That one wasn’t bad. Let’s see another one.

Next example (click here):

                                                            (2x+1)(4x+3y+2)

1. Distribute 2x+1

                                      (2x+1)(4x+3y+2)=2x(4x+3y+2)+1(4x+3y+2)
We distributed 2x+1 to make the problem easier for us to solve.
2. Use the distributive law

                                              2x(4x+3y+2)=2x∙4x+2x∙3y+2x∙2

                                                  1(4x+3y+2)=4x∙1+3y∙1+2∙1
We plug these values back in
                              (2x+1)(4x+3y+2)=2x∙4x+2x∙3y+2x∙2+4x∙1+3y∙1+2∙1

3. Simplify

                                         (2x+1)(4x+3y+2)=8x^2+6xy+4x+4x+3y+2
                                                           =8x^2+6xy+8x+3y+2

Last example (click here):

                                                                (x+1)(x+2)(x+3)

1. Apply the FOIL method to two of the polynomials

                                                      (x+1)(x+2)=x∙x+x∙2+1∙x+1∙2
                                                                     =x^2+3x+2

2. Plug x^2+3x+2 in for (x+1)(x+2)

                                                  (x+1)(x+2)(x+3)=(x^2+3x+2)(x+3)
Now this problem looks like the prior example.
3. Distribute x+3

                                           (x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
I always prefer to distribute the polynomial with the smallest terms, but you can pick to distribute the quadratic. In that case you would have:
                                           (x^2+3x+2)(x+3)=x^2 (x+3)+3x(x+3)+2(x+3)

4. Use the distributive law

                                            (x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
                                                   =x∙x^2+x∙3x+x∙2+3∙x^2+3∙3x+3∙2

5. Simplify

                                             (x+1)(x+2)(x+3)=x^3+3x^2+2x+3x^2+9x+6
                                                                    =x^3+6x^2+11x+6

Not bad! You can see when we break the problem down, step by step, it is a lot easier to solve because we use the basics of expanding (distributive law and FOIL). For more practice problems and help, check out our Practice.

Until next time,

Leah

Wednesday, October 11, 2017

Middle School Math Solutions – Expand Calculator, FOIL Method

In our last blog post we covered the distributive law. In this blog post, we will focus on an application of the distributive law, the FOIL method.

The FOIL method is used when multiplying two binomials together. Quick reminder: a binomial is an expression of the sum or difference of two terms.

The FOIL method is when we take the sum of the first two terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together. This is where we get the acronym FOIL (first, outer, inner last).

Let’s visualize this and see it in a formula:


Let’s see some examples now, using the FOIL method.

First example (click here):

                                                                 (x+3)(x+1)

1.  Use the FOIL method

Remember, we want the sum of the first terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together.
2.  Simplify

                                                     (x+3)(x+1)=x∙x+x∙1+3∙x+3∙1

                                                                  =x^2+x+3x+3

                                                                    =x^2+4x+3

   Multiply terms and add like terms

Not, too complicated. Let’s see two more examples.

Next example (click here):

                                                                    (2x+1)(x-1)

1. Use the FOIL method


2. Simplify

                                                        (2x+1)(x-1)=2x^2-2x+x-1

                                                                     =2x^2-x-1

Last example (click here):

                                                                 (3x^2+y)(-x+5y)

1. Use the FOIL method


2. Simplify

                                             (3x^2+y)(-x+5y)=-3x^3+15x^2 y-xy+5y^2

As you can see, multiplying binomials using the FOIL method isn’t hard. The FOIL method is so helpful throughout your math courses, so it is important to memorize and get the hang of it. Once you start practicing, this will become second nature to you. For practice problems or more help on the FOIL method, check out our Practice.

Until next time,

Leah

Monday, September 25, 2017

Middle School Math Solutions – Expand Calculator, Distributive Law

The distributive law helps with multiplication problems by breaking down large numbers into smaller numbers. In this blog post, we will talk about the distributive law and how to use it.

The law says that multiplying a number by a group of numbers added together is the same as multiplying each separately. What does this mean? Here we can see it in a formula:


You can see that the multiplication of a has been distributed among the sum of b and c.

Let’s see some examples of how to use the distributive law.

First example (click here):

                                                                Expand 4(2x+5)

1. Define the values for a, b, c

                                                               a=4,   b=2x,   c=5
Refer to the formula to see what these values are.
2. Plug these values into the distributive law formula

                                                               4(2x+5)=4∙2x+4∙5

3. Simplify your answer

                                                                4∙2x+4∙5=8x+20
8x+20 is our answer.
Next example (click here):

                                                                Expand 5(10-9p)

1. Define the values for a, b, c

                                                              a=5,   b=10,   c=-9p

2. Plug these values into the distributive law formula

                                                            5(10-9p)=5∙10+5∙(-9p)

3. Simplify your answer

                                                             5∙10+5∙(-9p)=50-45p
50-45p is the answer.
In this last example, we will see another application of the distributive law.

Last example (click here):

                                                                     5(106)

Just by looking at this problem, this might be difficult to calculate quickly. So let’s use the distributive law.

1. Make 106 the sum of two numbers

                                                                 106=100+6

2. Plug this in for 106

                                                                  5(100+6)

3. Define values a, b, c

                                                            a=5,   b=100,   c=6

4. Plug these into the distributive law formula

                                                           5(100+6)=5∙100+5∙6

5. Simplify your answer

                                                         5∙100+5∙6=500+30=530
530 is the answer.
As you can see, the distributive law is very handy. Therefore, I highly suggest you memorize this formula and become familiar with how to use it. You can do so by doing as many examples as you can. For practice examples and more help, check out Symbolab’s Practice

Until next time,

Leah

Tuesday, September 19, 2017

High School Solutions – Functions Calculator, Inverse

Last blog posts, we focused on how to find the domain and range of functions. In this blog post, we will discuss inverses of functions and how to find the inverse of a function.

An inverse of a function f(x), denoted f^(-1)(x), is a function that reverses or undoes f(x). What does that mean? This means that the domain or inputs of a function is the range or output of the function’s inverse and the range or outputs of a function is the domain or inputs of the function’s inverse. If f(x)=y, then f^(-1)(y)=x. Let’s see some pictures to better understand.

                                                         
It is important to note that some functions have more than one inverse. For example, quadratic equations have two inverses because the negative and positive value for an input goes to the same y value (For x^2+4, when x = 1 and x = -1 , we get y = 5). When you find the inverse of quadratics, you’ll notice you get two inverses, one is a postive square root and the other is a negative square root. We will see an example of this later in the post.

Now, that you’ve got the concept of what the inverse of a function is, we will see the steps on how to find the inverse of a function.

Steps to find the inverse of a function:

1. Replace y for f(x)
2. Solve for x
3. Substitute y = x
4. If you want to check the function, then f(f^(-1) (x))=x  and f^(-1) (f(x))=x

The steps are pretty simple to remember and follow. Now, let’s see some examples.

First example (click here):

                                                 Find the inverse of f(x)=3x+5

1. Replace y for f(x)

                                                                 y=3x+5

2. Solve for x

                                                                 y=3x+5

                                                                 y-5=3x

                                                               \frac{y-5}{3}=x

3. Substitute y = x

                                                       \frac{x-5}{3}=y=f^(-1) (x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

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

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

Next example (click here):

                                                     Find the inverse of f(x)=\sqrt{x+3}

1. Replace y for f(x)

                                                                     y=\sqrt{x+3}

2. Solve for x

                                                                     y=\sqrt{x+3}

                                                                     y^2=x+3

                                                                     y^2-3=x

3. Substitute y = x

                                                             x^2-3=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

                                               f^(-1)(\f(x))=(\sqrt{x+3})^2-3=x+3-3=x

                                              f(f^(-1)(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}=x

Last example (click here):

                                                   Find the inverse of f(x)=2x^2-2

1. Replace y for f(x)

                                                              y=2x^2-2                      

2. Solve for x

                                                                 y+2=2x^2

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

                                                              ±\sqrt{\frac{y+2}{2}}=x

3. Substitue y = x

                                                      ±\sqrt{\frac{x+2}{2}}=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)
You can do this step on your own. We will skip it to save some time.

As you can see, finding the inverse of a function is pretty simple. It is easy to make algebraic answers, so make sure you check your answer to see if it is correct. With practice, you’ll be able to master this topic easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Tuesday, August 15, 2017

High School Solutions – Functions Calculator, Parity (Even or Odd)

In this blog post, we will be discussing the parity of functions and how to find out the parity of a function.

The parity of a function is the attribute of being even, odd, or neither. A function is even, if \f(-x)=\f(x) for all x. A function is odd if \f(-x)=-\f(x) for all x. A function is neither even nor odd, when it satisfies neither of these options.

In order to find the parity of a function, we must see whether the statements about even and odd functions is true or false.

Steps to find the parity of a function:
  1.  Find \f(-x)
  2.  Find -\f(x)
  3.  See if \f(-x)=\f(x), \f(-x)=-\f(x), or neither
Simple enough! Let’s move onto some examples.

First example (click here):

                                                 Find the parity of \f(x)=x^2+4

1. Find \f(-x)

                                                         \f(-x)=(-x)^2+4

                                                               \f(-x)=x^2+4

2. Find -\f(x)

                                                      -\f(x)=-(x^2+4)=-x^2-4

3.  See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

                                                         \f(x)=x^2+4=\f(-x)

                                                  \f(-x)=x^2+4≠-x^2-4=-\f(x)

             \f(x) is even.

Next example (click here):

                                                   Find the parity of \f(x)=3x

1. Find f(-x)

                                                          \f(-x)=3(-x)=-3x

2. Find -\f(x)

                                                          -\f(x)=-(3x)=-3x

3. See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

                                                        \f(-x)=-3x≠3x=\f(x)

                                                          \f(-x)=-3x=-\f(x)

\f(x) is odd.

Last example (click here):

                                             Find the parity of \f(x)=\frac{x+2}{x+1}

1. Find \f(-x)

                                                        \f(-x)=\frac{-x+2}{-x+1}

2. Find -\f(x)

                                             -\f(x)=-(\frac{x+2}{x+1})=\frac{-x-2}{-x-1}

3. See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

                                          \f(-x)=\frac{-x+2}{-x+1}≠\frac{x+2}{x+1}=\f(x)

                                         \f(-x)=\frac{-x+2}{-x+1}≠\frac{-x-2}{-x-1}=-\f(x)

\f(x) is neither odd nor even.

As you can see, finding the parity of a function is very simple. Just memorize the definitions of even and odd functions and you are good to go! For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah