Factoring binomials using the greatest common factor

One way to factor binomials is by searching for the greatest common factor.

Consider

(5x * 25)

In this case the greatest common factor is 5 and the phrase can be written as

5(x*5)

Multiplying a trinomial by a binomial

Multiplying a trinomial by a binomial is a lot like multiplying a binomial by a binomial. You multiply the first term by all the factors of the second term, then multiply the outer(last) term by all the factors of the second term, then simplify.

Consider the example:

(x+5) (5x² + 3x + 6)

First we multiply our first term (x) by every term in the trinomial (5x² + 3x + 6)

this gives us:

(5x³ + 3x² + 6x)

next we multiply our outer term (5) by every term in the trinomial (5x² + 3x + 6)

this gives us:
(25x² + 15x + 30)

so we have
(5x³ + 3x² + 6x) + (25x² + 15x + 30)

we can simplify by adding like terms to get:
(5x³ + 28x² + 21x + 30)

Multiplying a binomial by a binomial

The most common way to multiply binomials is what is called the FOIL method

First
Outer
Inner
Last

Let us look at an example

(x+4) (x+1)

These are both binomials, to multiply them we first multiply the first two terms to get x²

Then we still take the first x and multiply it by the outer number: 1, to get x.

So far we have
x² + x

Now we do the inner number: 4

4 times x is 4x

and finally the last number 4 times 1 is 4

So in total we have
x² + x + 4x + 4

which can be simplified by combining the like x terms to

x² + 5x + 4

Polynomials

Polynomials can be anything from a single number to a variable to a combination of numbers and variables

Monomials have one term.
Such as... 8x⁴ , 6 , or 2xy

Binomials have 2 terms which are not like.
Such as... 2wz - 4dt , 4x² - 3x , 4c - 2d

Trinomials have 3 terms which are not like.
Such as... 4bt - 5yu + 9o , 3x² - 2x + 9 , 5t + 7y - 8u

Dividing exponents

Yesterday we learned that when you multiply exponents you add the number in the exponent, today we see that when you divide exponent you subtract the number in the exponent.

Consider

x⁷ / ³

What is this equal to?

x*x*x*x*x*x*x / x*x*x = x*x*x*x or x⁴

x⁷ / ³ = x^7-3 = x⁴

what about

x³ / ⁷

= 1 / x^7-3 = 1 / x⁴

We take the reciprocal because the exponent is greater in the divisor, or denominator.

Exponent

Exponents tell you how many times a factor is multiplied.

x * x * x (x times x times x)

Can be written as x³ or x^3 , when we write the multiplication in this way, we call it an exponent.

To multiply exponents we add them, for example, consider we have

x² * x³

what is this equal to?

x⁵

why is this? Well if we write it out, it becomes obvious


x² * x³

=


(x*x) * (x*x*x) or x⁵

if we have

5x² * 2x³

Then the bottom numbers (or base numbers) are multiplied, while the exponents are added


5x² * 2x³
=
10x⁵

Using Substitution to Solve a System of Equations

Suppose we had a system of equations

2x + y = 4
and
3x + 2y = 5

How can we solve for x and y?

The good thing is that we have two equations for two variables.

One way is to solve on equation for y and substitute. Let us start with

2x + y = 4

subtract 2x from both sides

y = 4 - 2x we can use this informaiton to solve for x by substituting y into the other equation

3x + 2y = 5 becomes

3x + 2(4-2x) = 5

3x + 8 - 4x = 5

-x = -3
x=3

so x = 3

Now we can substitute x into our first equation to find y

2x + y = 4

6 + y = 4

y = -2

To check let us substitute our answers into the equations and see if we get the same answer:


2x + y = 4
and
3x + 2y = 5

x=3 y= -2

2(3) - 2 = 4
6-2 =4 Correct.

Next

3(3) + 2(-2) =5
9 - 4 = 5 Correct.

So our solutions to the system check OK and are correct.