# Unit 4 Review

## Overview

Learning how to factor quadratic equations is very important, because it makes solving quadratic equations much easier. You will have to factor and solve in many other units, so it is important to have a strong base on what we are going to learn today. This sheet will help you discover the many different methods and approaches available for you to factor and solve quadratic equations.

## Distributive Property

What is Distributive Property?

It is an algebra property which is used to multiply a single term and two or more terms inside a set of parentheses.

Multiplying a Monomial by a Polynomial

1) (3)(x + 2)

You must multiply 3 by the terms inside the brackets. Your new expression will be...

3x+6

2) (2x)(1 + 5x - 3x^3 )

Multiply 2x by the terms in the brackets. Your new expression will be...

2x+10x^2-6x^4

**note: when you multiply terms with exponents, you must add their powers. For example, 2x * 2x will be 4x^2

Multiplying Binomials

Remember FOIL .(F stands for first, O for outer, I for inner, and L for last)

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

Multiply x (in the first binomial) by the first term in the second binomial, which is x. Then, multiply x (in the first binomial) by the outer number of the parentheses. Next, the inner terms of the expression (2 multiplied by x). Finally, multiply 2 (in the first binomial) by the last term of the expression (3). Your new expression will be....

x2 + 3x + 2x + 6

Now, we can simplify this. Remember simplifying means collecting like terms so you have a more concise and simple espression. So in this case, we can add 3x and 2x, because these numbers are like terms (terms that contain the same variable and raised to same power). This expression simplified will be...

x2 + 5x + 6

## Common Factoring

What is Common Factoring?

This method is important to know and use when the quadratic equations get more and more complicated. Common factoring is a process used to find all the factors that are multiplied to give a polynomial expression. Common factoring is the opposite of distributive property, where instead of expanding the brackets, you are factoring the terms in parentheses.

Factoring Polynomials

1) 7x - 14

In order to factor this polynomial, you must figure out a factor that can be divided by the terms in your expression. In this case, 7 can be factored in all of your terms in this expression. We will have to place 7 out of the brackets, and inside the brackets will be x and -2. Your solution is...

7(x-2)

Now, when you open the brackets using the distributive property, it will give you 7x-14 (the original expression).

2) 4x^3 - 9x^2

The common factor in this polynomial will be x^2, because x is found in both terms, and the power of 2 is also in both terms. So we will place x^2 out of the brackets, and we will place the rest inside of the brackets, like so...

x^2(4x-9)

When you use distributive property again, you will get the original expression.

3) 4x^4 + 6x^3 - 8x^2 + 10x

The common factor is 2x, and this factor can be divided be all of these terms. Now place 2x outside of the brackets, and put the remainder of the expression inside the brackets. Your new expression will be...

2x(4x^3+6x^2-8x+10)

Factoring by Grouping

2x^2+4x+3x+12

Since there are four terms, we are able to factor them into binomials. First, we must find the common factor between 2x^2 and 4x, which is 2x, and place it outside of the brackets. Then we place the remainder inside of the brackets.

2x(x+2)+3x+6

Next, we will work on the remaining two terms, 3x and 12. The common factor between these two is 3, and we place it outside of the brackets. Now, we put the rest inside of the brackets. This should give you a solution of...

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

What repeats in this equation? It is x+2, so we can simply write x+2 as one binomial, and then write the remainder of the equation, which will be 2x+3. Our polynomial factored will be...

(2x+3)(x+2)

## Factoring Simple Trinomials (Quadratic Relations)

What is Factoring Simple Trinomials?

Factoring simple trinomials deals with a polyomial with three terms. The polynomial should be in the form of x^2+bx+c+y (there is no a value) to factorize using this method.

1) x^2+ 7x + 12

In order to factorize this trinomial, you must find two numbers that will give you the sum of your b value (7) and your c value (12). In this case, the two numbers will be 4x and 7x. Rewrite your polynomial like so:

x^2+4x+3x+12

Now, we will factor by grouping, like we have learned previously.

x(x+4)+3(x+4)

We can simplify this even further, by looking at what is repeating in this polynomial. In this case, it is x+4. We can write it as one, and then write the rest of the terms.

(x+4)(x+3)

## Factoring Complex Trinomials

What is Factoring Complex Trinomials?

Factoring simple trinomials deals with a polyomial with three terms. The polynomial should be in the form of ax^2+bx+c+y (there should be an "a" value other than 1) to factorize using this method.

1) 3x^2 - 10x+ 8

Unlike simple trinomials, there is an 'a' value, which is 3 in this situation. We must consider the a value when we are dealing with complex trinomials, so we are going to multiply the a value by the c value, which will give us 24. With the c value now being 24, and the b value being -10, we are going to figure out the two numbers which will give us the sum and product (like we have learned in simple trinomials)

3x^2-6x-4x+8

Now, we are going to factor by grouping.

3x(x-2)-4(x-2)

Our factored form is....

(3x-4)(x-2)

## Difference of Squares

What is a Difference of Squares?

It is when a squared number is subtracted by another squared number. These terms MUSTbe subtracted from one another, or else it would not be a difference of squares.

1) x^2 - 16

The square root of x^2 is x, and the square root of 16 is 4. We will rewrite this as...

(x-4)(x+4)

**one of the binomials must have an addition symbol between the terms, and the other binomial must have a subtraction symbol. When you expand this, you will get the original form (x^2-16)

2) 100 - x^2y^2

The square roots of each term in the binomial is 10 and xy. We will write this a difference f squares, like so:

(10-xy)(10+xy)

3) 49x^6 - 36y^4

The squareroots of this is 7x^3 and 6y^2. We will rewrite this as...

(7x^3-6y^4)(7x^3+6y^4)

The quadratic equation can be described as: ax^2+bx+c=y. A quadratic equation is a non-linear equation, meaning it does not follow a straight line, but like a curve - a parabola. Unlike a linear equation, the quadratic equation has second differences instead of a first difference (where the differences are constant between the terms)

You can solve a quadratic equation by solving for x (AKA roots). There are three ways for you to solve for your roots: solving for x in factored form, completing the square, and using your quadratic formula.

Solving for X in Factored Form

x^2+6x-16=y

Factor this by using the method for simple trinomials.

(x+8)(x-2)=y

We know that an x intercept (a root) has the coordinates of (x,0), so we have to we to figure out a number that we can sub in for x that will give us 0 (for y). In this case it is -8 and 2 (-8+8=0 and 2-2=0). As a result, our roots are x= -8, 2.

**remember that a parabola has two roots, so therefore there will be two answers as your solution.

Completing the Square

x^2+2x-3=0

We are going to isolate the x by using this method. Our first step is to block off the first two terms, and place the a value outside of the brackets.

(x^2+2x)-3=0

Now, we are going to add our zeroes. We need another number to place inside of the brackets so it becomes into a trinomial, and from there we are able to factor simple trinomials. This zero can be found by dividing the c value by 2, and then squaring it. Our zero will be 1, but we will also need to place a '-1' so it doesn't alter the equation and balance it out.

(x^2+2x+1-1)-3=0

Place '-1' outside of the brackets.

(x^2+2x+1)-3-1=0

Now, we are going to factor by simple trinomials.

(x+1)(x+1)-4=0

(x+1)^2=4

Square root both sides, so you can get rid of the square.

x+1=2

x=-3, 2

Using the quadratic formula below, we will be able to find our roots. All we have to do is sub in our a, b, and c values into the equation, and solve.

x^2+x-6=0

a= 1

b= 1

c= -6

x=-1+-[sqrt(1^2-4(1](-6)]/2(1)

x=-1+-[sqrt(1+24)]/2

x= -1+5/2, -1-5/2

x= 2, -3

## Unit 4 Summary

With the many different methods of factoring and solving for roots, we are now able to gather up all of our knowledge to factor and solving various problems. Here are some challenging questions we are going to tackle:

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

(x-3) repeats several times. In order to make factoring this expression much more easier, we will let 'a' represent x-3. Let's write the reminder of the expression, like so:

a^2+3a-4

Now, let's factor simple trinomials...

(a+4)(a-1)

We can sub in 'x-3' for 'a'...

[(x-3)+4][(x-3)-1]

Finally, let's simplify this...

(x+1)(x-4)

2) Factor 16x^4-1

First, let's factor this by using the difference of squares method...

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

We are able to factor this even further, because there is another difference of square, which is 4x^2-1. This becomes into...

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

3) Solve 2x+1= [square root(2x+1)]

Get rid of the square root.

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

4x^2+2x+2x+1=2x+1

4x^2+2x=0

4(x^2+1/2x)=0

4(x^2+1/2x+1/16)-1/4=0

4(x+1/4)^2-1/4=0

2(x+1/4)=1/2

(x+1/4)=1/4

x=0, -1/2

4) 5x=180/4x

Get rid of the fraction by multiplying 4x by both sides.

20x^2=180

20x^2-180=0

20(x^2-9)=0

Factor by the difference of squares method.

20(x-3)^2=0

Isolate x.

(x-3)^2=0

x-3=0

x= 3, -3