# Unit 4 - Review Package

### By Matthew Huynh

## Welcome

Welcome to my math review package. Here, you should be able to review the Quadratic Functions math unit (Unit 4). Skills that will be reviewed are expanding, quadratic factoring, and solving for a variable using various different methods.

## Identifying a Quadratic Relation

To begin, we should remember how to identify a quadratic relation. We can figure this out by plotting a relation on a table of values. If the independent variable (x) was to have constant increments, then the second differences of the dependent variable (y) should be equal. Quadratic relations will always have an **equal second difference** for the y-value.

You can also check for the degree of the fully expanded equation. The degree for a fully expanded quadratic equation with polynomials should be **two**.

4x^2 – 2x + 4

This equation has a degree of two.

## Multiplying Binomials

“Use the distributive property to expand binomials. Collect like terms to simplify.” Being able to multiply binomials correctly is important because this is the process of **expanding**. Expanded expressions should also be **fully simplified.**

(x + 2)(x+5) We can use the FOIL method to multiplying these two binomials

= x^2 + 5x + 2x + 10 Now we collect like terms

=x^2 + 7x + 10 Done

Like the diagram shows, the Firsts multiply with each other. The Lasts multiply with each other. The Outers multiply with each other. And the Inners multiply with each other.

## Common Factoring

Factoring is an important skill as it will be used all throughout the quadratics unit. Factoring is the **opposite of expanding**. Basically, factoring is when you **take out the greatest common factor** (GCF). To determine a common factor, you must look at each term of the polynomial and make sure it is divisible by that factor. A polynomial is not completely factored until the GCF has been factored out.

= 4x + 20 Factor this. The common factors are 1, 2, and 4. We will use 4 as it’s the GCF.

= 4(x+5) The 4 goes outside the bracket. Inside the bracket is x + 5. If you expand this it would give 4x + 20

## Factoring by Grouping

Sometimes, expressions come with **many terms.** It is easier to **create groups** of terms and factor them individually. Basically, this is breaking the expression into **two or more parts** and factoring each individual part.

= x^3 + 2x^2 + 8x + 16 This is the equation to be factored. We can break the equation in half. We can find the factors of each (x^3 + 2x^2) and (8x + 16)

= x^2(x+2) + 8(x+2) x^2 is factored out from the first and 8 is factored from the second.

= (x+2)(x^2 +8) Since we are left with both as (x+2), we can collect like terms.

You can click on the picture to expand it.

## Factoring Simple Trinomials - y=Ax^(2 )+ Bx+C

It is necessary to factor using this method if you cannot see any common factors between the terms. An example of this would be if one of the terms were a **prime number** such as 1 or 7. It is still possible to find common factors from expressions with prime numbers.

Let's look at the trinomial: x2 + 7x + 6

We'll label x2 as the first term, 7x as the second term, and 6 as the third term.

There are two prime numbers in this expression, 1 and 7. There are not obvious factors that can be seen so we must use a different method. We must find two numbers with a **product** equaling the third term and a **sum** equaling the second term.

x2 + 7x + 6 Two numbers with a product of 6 and sum of 7? 6 and 1.

After you have found the two numbers. You can put them into factored form.

(x+r)(x+s) Where r is the a number and s is the other number. It does not matter which of (x+1)(x+6) the two numbers are r or s.

We can check this by expanding and simplifying

= (x+1)(x+6)

= x2 + 7x + 6

## Factoring Complex Trinomials by Decomposition y=a(x-r)(x-s)

Complex trinomials are **different from simple trinomials** because you may not be able to find the two product and sum numbers.

5x2 + 12x + 4 There are no numbers that give a product of 4 and sum of 12.

When this happens, we must **factor by decomposition**. This means that the middle term is decomposed (split) into two numbers.

To factor by decomposition, we **take the A term and multiply it by the C term**. Then we can try and find two numbers that give a product of this new number and a sum equal to the B term.

5x2 + 12x + 4 Multiply 5 by 4. The product is 20.

Now we find two numbers with a product of 20 and sum of 12.

The numbers are 2 and 10.

After finding the numbers, we **decompose** the middle term.

5x2 + 10x + 2x + 4 The values 10 and 2 were taken from above. Notice how 10x + 2x is equal to 12x.

5x2 + 10x + 2x + 4 We can now factor this expression by grouping.

5x(x+2) + 2(x+2)

(5x + 2)(x + 2) Done.

## Factoring Difference of Squares

First we should know how to identify a difference of squares.

x2 - 4

This expression is a difference of squares. Notice how the C term is negative. This term **must be negative** for the expression to be a difference of squares. The C term can also be **square rooted** giving a whole number. You **must be able to square root any C term** of a difference of squares.

To factor this expression, we square root the C term

= x2 - 4

=(x+2)(x-2)

We can check this by expanding

=(x+2)(x-2)

= x2 - 4

## Solving a Quadratic Equation

There are various methods that can be used to solve an equation. Recall that you must isolate a variable for an equation to be solved. There are various methods we can use to solve for variables however we will focus on factoring the equation, and by using the quadratic formula.

## Solving from Factored Form

If the equation is in factored form we can use this method. Otherwise, we can try converting it to factored form by fully factoring.

x2 + x + 2 = 0 Factor.

(x+1)(x+2) = 0

From here, we can set either (x+1) or (x+2) to 0 to solve for x. The reason why we set either one to 0 is because when both (x+1) and (x+2) multiply, the product is 0.

x+1 = 0

x = -1

x+2 = 0

x = -2

x = -1,-2 There can be two solutions, as x can be both -1 and -2. We can check this by plugging both back into the equation.

## The Quadratic Formula

## Completing the Square

Completing a square is done to convert a quadratic equation from Standard Form to** Vertex Form**, which should include a **perfect square**. There are a few reasons to convert an equation to vertex form: it will be **easier to graph the equation**, and you can still solve for x.

y = (x – h)^2 + k This is the vertex form

Here are the steps to complete a square:

y = 4x^2 +16 x – 2 Here is an equation in standard form

y = (4x^2 + 16x) - 2 We will first start by blocking off the first two terms

y = 4(x^2 +4x) – 2 Now we factor out A from the standard form. x^2 must be by itself

y = 4(x^2 + 4x + 4 – 4) – 2 We now have to add “zero”. The value for zero is found by dividing B by two, then squaring it. (4/2)^2 = 4. In this equation, it is written as + 4 – 4 which is equal to 0.

y = 4(x^2 + 4x + 4) -2 -16 We have to bring out the negative “zero” value but remember, since we’re taking it out of the bracket, we remember we still need to follow the Distributive Law. That is why the -4 turns into + 16

y = 4(x^2 + 4x + 4) – 18 Now we have to factor whatever is left inside the brackets

y = 4(x + 2)^2 - 18 Done! Notice how there is a perfect square.

To find the vertex of a parabola given an equation in vertex form, take the h and k values and substitute it into (h, k) where h is the x variable and k is the y variable.

y = (x – h)^2 + k Remember, this is the formula for vertex form.

In the previous equation above, the vertex would be (-2, -18). +2 turned into -2 since vertex form has a negative sign (-) before the h indicating that the h in the form is negative.

## Solving a Word Problem (1) Here is an example of a quadratic word problem that is solved by completing the square and using what we know about the vertex form | ## Solving a Word Problem (2) | ## Solving a Word Problem (3) |