Quadratics Review | Completing the Square to get Vertex Form |

Factoring Methods | Obtaining Quadratic Equations from a Graph or Points |

Completing the Square (Square Root Method) | More Practice |

Note that factoring **the sum and** **difference of cubes**, and more advanced **polynomial** **factoring** and **exponential** **factoring** can be found in the **Advanced Factoring** section.

Again, there are several ways to solve quadratics, or find the solution (also known as the $ x$-intercepts, roots, zeros, or values) to a quadratic equation.

In the **Introduction to Quadratics** section, we learned two ways to solve quadratics:

**Quadratic Formula**: For $ 0=a{{x}^{2}}+bx+c$, solutions are $ \displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$. This method can always be used!**Graphing Quadratics**: Looking for zeros by pinpointing where the quadratic graph ($ y=a{{x}^{2}}+bx+c$) crosses the $ x$-axis.

In this section, we’ll talk about two other ways:

**Factoring and setting factors to 0.**However, not every quadratic can be factored.**Completing the square and taking the square root of each side**. We’ll also use this to go from**Standard Form**to**Vertex Form**.

Before we get into the details, here is a Quadratic Review that might be helpful:

## Quadratics Review

Let’s review the different** forms of Quadratics** and also the **methods for finding roots** of Quadratics. **Note that the “**$ \boldsymbol{a}$**” (coefficient of the **$ \boldsymbol{{{{x}^{2}}}}$**) is the same for all three forms!**

**Note**: Sometimes we’ll have a Quadratic in “almost” **vertex form** and play around with it to get it in vertex form. For example, if we have $ \displaystyle y=-8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3$ and want to change it to vertex form, we can take out the $ \displaystyle \frac{1}{2}$ coefficient of $ x$ and do some algebra: $ \displaystyle \begin{align}y&=-8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3=-8{{\left( {\left( {\frac{1}{2}} \right)\left( {x+4} \right)} \right)}^{2}}+3\\&=-8{{\left( {\frac{1}{2}} \right)}^{2}}{{\left( {x+4} \right)}^{2}}+3=-2{{\left( {x+4} \right)}^{2}}+3\end{align}$, so the vertex is $ (–4, 3)$.

**Note: **If we are really having a difficult time factoring a quadratic trinomial, we could, as a last resort, use the **Quadratic Formula**, get the roots, and, if they are rational (integers or fractions), **put the quadratic back in factored form**. For example, if we used the Quadratic Formula and got roots $ \displaystyle \frac{4}{7}$ and **–3**, our factors would be $ \left( {7x+4} \right)\left( {x-3} \right)$.

Now we’ll get into the nitty-gritty:

## Factoring Methods

Just the way we learned how to multiply binomials (via **FOILING**), we need to learn how to do the opposite, or **factor** (or “**unfoil**”) the resulting trinomials. After we’ve done enough multiplying binomials and factoring trinomials, this will become second nature. We’ll also learn other basic polynomial factoring methods, like taking out the **Greatest Common Factors (GCF) **of polynomials, and factoring the **difference of two squares** and factoring **perfect square trinomials**. Think of factoring as just “pulling apart” things that are multiplied together. It’s the same principle of factoring **35** and getting **5** and **7**. When we factor in algebra, we do the same thing, but with variables.

Here are some important notes about using factoring to solve:

- When we factor to solve, we set each factor with a variable in it to
**0**, and solve for the variable to get the roots. This is because any factor that becomes**0**makes the whole expression**0**. (This is the zero product property: if $ ab=0$, than $ a=0$ and/or $ b=0$). - When we factor to solve quadratics or any polynomials, we can never just divide by factors (with variables) on both sides to get rid of them. If we do this, we may be missing solutions!
**If the coefficient of the term with the**$ \boldsymbol {{{x}^{2}}}$**is negative, I like to factor it out and just add a negative sign in front of the factors after factoring.**- Not every quadratic can be factored; if it can’t, we’ll have to use one of the other methods, such as the Quadratic Formula.

Let’s first learn **factoring techniques** so we can solve the quadratics that can be factored; it’s a pretty simple way to solve them. And later, when we learn how to solve more generic polynomials in the **Graphing and Finding Roots of Polynomial Functions** and **Advanced Factoring** sections, we’ll know how to factor!

**HINT**: In many of the examples below, we’ll need to find factors of a number to see which ones might work in factoring a quadratic. We can use a **Graphing Calculator** to help with this: Press **y=**, enter the number to be factored/x (for example, $ \frac{{144}}{x}$), press **2 ^{nd} graph (table)**, and the factor pairs will be displayed (look for the integer ones). You may want to go to

**2**

^{nd}**window (tblset)**and set

**TblStart**to

**0**, and $ \Delta $ tbl to

**1**before that.

**Taking out the Greatest Common Factor (GCF)**

Let’s first start with polynomials that need some simple factoring. We always need to pull out the **GCF** (**largest coefficients/variables that go into all the terms**) first. (We first learned about the **GCF** with regular numbers here in the **Multiplying and Dividing Section**.)

Remember from the **Exponents and Radicals in Algebra** section that we add exponents when we multiply with the same base, and subtract exponents when we divide with the same base.

Here are some examples of factoring polynomials by taking the **GCF** out (which is actually dividing!). Remember to always **multiply back to check the factoring**; it’s much easier to multiply than divide.

See how it’s reversing the distributive property? It’s “undistributing”.

**Factoring Trinomials (Quadratics)**

When we factor quadratics, we try to “UNFOIL” to get two binomials; we learned how to “**FOIL**” here in the** Introduction to** **Multiplying Polynomials** section. **Again, remember that when factoring trinomials, we always need to take out any GCF’s first!**

**This method does not always work; certain quadratics are “unfactorable” (prime, or irreducible), and must be solved with another method. **Remember that, with a trinomial, if the discriminant $ {{b}^{2}}-4ac$, where $ a$, $ b$, and $ c$ are from $ a{{x}^{2}}+bx+c$, isn’t a perfect square (like **25** or **49**), you can’t factor it. Here are some methods that we use for factoring trinomials:

#### Special Products of Binomials – Easily Factored

There are two special cases where you can use a shortcut to factor the trinomials: **Difference of Two Squares** (actually not a trinomial, but a binomial!), and **Perfect Square Trinomials**. We saw these here in the** Introduction to Multiplying Polynomials** section. Here are the two cases and how to factor them:

**Guess and Check**

“**Guess and Check**” is just what it sounds; we have certain rules, but we try combinations to see what will work.

**NOTE**: Always take a quick look to see if the trinomial is a **perfect square trinomial**, but you try the guess and check. In these cases, the middle term will be twice the product of the respective square roots of the first and last terms, as we saw above. For example, $ \displaystyle {{x}^{2}}-12xy+36{{y}^{2}}=\left( {x-6y} \right)\left( {x-6y} \right)={{\left( {x-6y} \right)}^{2}}$.

Let’s start with an example; let’s “**UNFOIL**” or factor one of the first examples that we worked with in **Quadratics**. Here’s how we **FOIL**’ed, and how we “**UNFOIL**“:

Note the **signs** when you “unfoil”; multiply these back to make sure they work:

Sometimes it’s easier and more visual if you use an “**X**” when you begin to factor these; we’ll first show this when there are no numbers (coefficients) before the $ {{x}^{2}}$ term. We can put the coefficient of the constant term (“$ c$”) on the top, middle term (“$ b$”) on the bottom, and then use the two sides to find two numbers **multiplied together to equal the bottom**, but **added together to equal the top**. This is sometimes called the “**A and M**” or “**AM**” method, since we are looking for numbers that both add and multiply to certain numbers.

Let’s try this for the last problem above: $ {{x}^{2}}-3x-10=\underline{{\left( {x-5} \right)\left( {x+2} \right)}}$

Let’s try another one that’s a little bit more complicated. **Note again that to make the factoring easier, we always want to factor the GCF out of the polynomial before we “unfoil”.**

**Example: **Factor $ 8{{x}^{2}}-22x+14$. First we take out the GCF (**2**) to get $ 2(4{{x}^{2}}-11x+7)$. We still need to “unfoil” $ 4{{x}^{2}}-11x+7$. This is a little more complicated, since we have the $ 4{{x}^{2}}$. Now we need **two factors of 4** and **two factors of 7**, so that **multiplying the inside terms** and **multiplying the outside terms** and **adding them together** will give us **–11**. (We can see that this trinomial is factorable, since $ {{b}^{2}}-4ac$, the discriminant, is $ {{11}^{2}}-4\left( 4 \right)\left( 7 \right)=9$, which is a perfect square.)

When we’re “guessing and checking”, we’re actually “foiling back” to see if we got the right answer! You may want to use your factor trees (that we learned in the here** in the Multiplying and Dividing **section) to write down all the factors of the first and last coefficients. Here are all the combinations we could have. Note that the factors of **4** are **2** and **2**, or **4** and **1**. The factors of **7** are just **7** and **1**.

The complete factoring is**: **$ 2(4x-7)(x-1)$. If we were to solve this trinomial, we would get $ \displaystyle \left\{ {x:\text{ }x=\frac{7}{4},\text{ }1} \right\}$ by setting each factor to **0** and solving for $ x$*. *Ignore the factor of **2**, since **2** can never be **0**; if we had an $ x$ on the outside, an additional factor would be **0**.

Note that if we had $ 2(4{{x}^{2}}-11xy+7{{y}^{2}})$, we’d also have to split the $ y$’s at the end, so we would factor to $ 2(4x-7y)(x-y)$. Multiply it all to together to show that it works! (This is one we wouldn’t typically be asked to solve).

**Grouping Method, or the “AC” Method**

The **Grouping Method** for factoring trinomials has gotten more popular since the FOILing hasn’t been taught as much recently. (The reason it’s not taught as much is because it can be only used with binomial multiplication. I still like FOILing since I believe learning to factor is easier if we learn FOILing first).

With the **AC method**, we will first turn the trinomial into a quadratic with four terms, to be able to do the grouping. Then, we have to find a pattern of binomials so we can use the distributive property to put them together (like a puzzle!). Let’s look at the same problem as above (with the **2** already factored out, but let’s keep the $ y$’s at the end):

$ 4{{x}^{2}}-11xy+7{{y}^{2}}$

Let’s use our “**X**” again to help us solve by the grouping method. We can put the middle term’s coefficient “$ b$” (**–11**) on the bottom part of the “**X**”, but this time we **multiply the first and last coefficients **(the “$ a$” and “$ c$” of $ a{{x}^{2}}+bx+c$) and put it on the top part of an “**X**”: $ 4\times 7=28$. Remember that the **sign of a term comes before it**, and pay attention to signs.

Then we do the “guessing and checking” as follows:

Since we found that **–4** and **–7** “work”, we can rewrite the trinomial and **separate the middle term**, so we can factor with the grouping method:

### Grouping Method with Four Terms

Here are some examples of the grouping method when we start out with four terms – which is actually easier! Notice that the first one is a four-term quadratic and the second is a cubic polynomial that includes factoring with the **difference of squares**. Remember the **difference of squares** factoring is $ \displaystyle {{x}^{2}}-{{y}^{2}}=\left( {x-y} \right)\left( {x+y} \right)$.

Note that having four terms to factor doesn’t always work with the grouping method above; sometimes we have to look for **differences of squares** (sometimes combined with the grouping method). These are tricky, but you might see them!

### The Box Method

One more method that is getting popular is called the “**Box**” method. This is a modified “**AC**” method, but we use greatest common factors (**GCF**) to help us factor. This is sort of the opposite of what we did with the **Multiplication Box** in the **Introduction to Multiplying Polynomial** section.

Using a **2** by **2** box, put the first term in the upper left hand corner, the last term in the lower right hand corner. Then **divide up the middle term in the remaining two corners so they add up to the middle term**, and are** factors of the first term times the last term** (thus, the “AC” method). You can put the middle terms (upper right and lower left corners) in any order, but make sure the signs are correct so they add up to the middle term. If you have set up the box correctly, the diagonals should multiply to the same product.

Then get the **GCF**s across the columns and down the rows, using the **same** **sign of the closest box** (boxes either on the left or the top).

Try this for $ 10{{x}^{2}}-11x-6$:

Then read across and down to get the factors: $ (5x+2)(2x–3)$. Foil it back, and we see that we got it correct! If we were to solve this trinomial, we would get $ \displaystyle \left\{ {\,x:x=-\frac{2}{5},\,\,\,\frac{3}{2}\,} \right\}$ by setting each factor to **0** and solving for $ x$.

### The “Star” Method

Another method for factoring trinomials is the “**Star**” Method, which is kind of a combination of the Guess-and-Check “**X**” method and the “AC” method. Use $ 10{{x}^{2}}-11x-6$, setting up the “star” as follows:

### The “AC” Method without Grouping

There’s another new method to factor trinomials out there; it’s similar to the “star” method, but without drawing the “star”. Again, use $ 10{{x}^{2}}-11x-6$:

## Completing the Square (Square Root Method)

**Completing the square** is what is says: we take a quadratic expression in standard form $ a{{x}^{2}}+bx+c$ and manipulate it to have a **binomial square** in it, like $ a{{\left( {x+b} \right)}^{2}}+c$. This way we can solve it by isolating the binomial square (getting it on one side) and **taking the square root of each side**. This is commonly called the **square root method**. We can also complete the square to find the vertex more easily, since the vertex form is $ y=a{{\left( {x-h} \right)}^{2}}+k$ (see below).

First think about what happens when we square a binomial by looking at: $ \displaystyle \begin{align}{{\left( {x+3} \right)}^{2}}&=\left( {x+3} \right)\left( {x+3} \right)\\&={{x}^{2}}+3x+3x+9\\&={{x}^{2}}+6x+9\end{align}$. (We first saw these **perfect square trinomials** in the here in the** Introduction To Polynomials **section.) Since we add the product of the middle terms twice, we have twice the product of the first ($ x$) and second (**3**) terms in the middle (to get $ 6x$). See also how we have the square of the second term (**3**) at the end (**9**). Thus, to complete the square of a trinomial that isn’t a perfect square, we need to **halve the second term** and **square it** – and then **add that number so the square can be complete**. Then we have to make sure to **add the same thing to the other side.**

Then, we **take the square root of each side**; remember that we need to include **the plus and minus **of the right hand side, since by definition, the square root is just the positive. Another way to think of it is the **absolute value of the left side equals the right side**, so we have to include the plus and minus of the right side.

**(Note**: you may want to review solving square root equations here in the **Solving Exponential and Radical Equations **section.**)**

Let’s work with $ {{x}^{2}}+8x+15=0$ first, since it just starts with an $ {{x}^{2}}$ (coefficient of $ {{x}^{2}}$ is **1**):

Note that it would have been **much easier** **to factor** this quadratic, but, like the quadratic equation, we can use the completing the square method for **any quadratic**. (We may not always get a real number for the answer(s); we’ll talk about imaginary numbers later).

Here’s one that’s a little trickier, because of the radical in the coefficient of $ x$:

Here’s one where the coefficient of the $ {{x}^{2}}$ isn’t **1**. Remember again that if we can take out any factors (GCF) across the whole trinomial, do it first and complete the square with the trinomial only. This gets a little more complicated, but it’s not too bad:

**Note**: There is another method for **Completing the Square** from **Standard Form** to solve. For **standard equation** $ \displaystyle a{{x}^{2}}+bx-c=0$, first subtract $ c$ from both sides to get in the form $ \displaystyle a{{x}^{2}}+bx=-c$. If $ a$ is a perfect square, add $ \displaystyle \frac{{{{b}^{2}}}}{{4a}}$ to both sides and then factor and simplify. If $ a$ is not a perfect square, first multiply both sides by anything that makes $ a$ a perfect square (you can always multiply by $ a$), add $ \displaystyle \frac{{{{b}^{2}}}}{{4a}}$ to both sides and then factor and simplify. Try this for the example: $ \displaystyle 2{{x}^{2}}-5x-12=0$: $ a=2$, $ b=-5$, and $ c=-12$:

I’m not a fan of these “memorization methods” since I’d rather understand what I’m doing! But if it’s easier for you, go for it!

**Note**: Even though **complex solutions** are easier to obtain using the **Quadratic Formula**, here are examples on how we can get imaginary solutions by **Completing the Square**:

## Completing the Square to get Vertex Form

We can use the same technique to put $ y=2{{x}^{2}}-5x-12$ (standard form, or $ y=a{{x}^{2}}+bx+c$) into vertex form ($ y=a{{(x-h)}^{2}}+k$). In this case, we want to leave everything on **one side**, so we have to undo what we’ve added or subtracted by completing the square:

**Note:** There is another way to convert from **Standard Form** to **Vertex Form**. For **standard equation** $ \displaystyle y=a{{x}^{2}}+bx+c$, we can use a **General Vertex Form** equation $ \displaystyle y=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}-\frac{{{{b}^{2}}-4ac}}{{4a}}$ to convert to **vertex form***. *Try this for the example $ \displaystyle 2{{x}^{2}}-5x-12=0$: $ a=2$, $ b=-5$, and $ c=-12$:

$ \displaystyle \begin{align}y&=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}-\frac{{{{b}^{2}}-4ac}}{{4a}}\\&=2{{\left( {x+\frac{{-5}}{{2\left( 2 \right)}}} \right)}^{2}}-\frac{{{{{\left( {-5} \right)}}^{2}}-4\left( 2 \right)\left( {-12} \right)}}{{4\left( 2 \right)}}\\&=2{{\left( {x-\frac{5}{4}} \right)}^{2}}-\frac{{121}}{8}\,\,\end{align}$

Again, I’m not a fan of these “memorization methods” since I’d rather understand what I’m doing!

## Obtaining Quadratic Equations from a Graph or Points

Note that we will learn more about parent graphs and transformations in the **Parent Functions and Transformations** section.

Sometimes you will be asked to look at a quadratic graph (or given the vertex and a point) and write the equation, in any form, for that graph. The easiest way to do this is to find the vertex from the graph, if possible, put the equation in **vertex form**, and then **compute the “**$ \boldsymbol {a}$**”** (coefficient of $ {{x}^{2}}$) from another point on the graph, such as a root or $ y$-intercept.

If you’re given the $ x$-intercepts (roots), you can also put the quadratic in **factored form**, and use another point, which is not one of the roots, to find the “$ a$” part of the equation. Remember that the “$ a$” in all three forms (standard, factored, and vertex) will be the same.

Here are some examples:

Here’s another type of problem you might see where you have to **write a Quadratic Function** given a **Parabola’s Axis of Symmetry and two Non-Vertex Points**. Note that we have to use a **System of Equations**:

Here’s one more where we are given three points and need to **find the equation of the parabola**. Note that we could have also used the calculator to perform a **quadratic regression** with the points, as we did here in the **Scatterplots, Correlation, and Regression** section.

**Learn these rules, and practice, practice, practice!**

**For Practice**: Use the **Mathway** widget below to try a **Factoring** problem. Click on **Submit** (the blue arrow to the right of the problem) and click on **Factor** to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on **Tap to view steps**, or **Click Here**, you can register at **Mathway** for a **free trial**, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to **Quadratic Inequalities** – you are ready!