Factoring Quadratic Equations - Examples & Practice

(Video) Solving Quadratic Equations by Factoring

by PatrickJMT

This video by PatrickJMT covers quadratic equations.

$\large{\text{Summary}}$

We combine factoring and the zero product property to solve quadratic equations. The general steps to solving a quadratic equation are as follows:

1. Manipulate the equation so you have a quadratic set equal to $0$.
   1. If possible, remove common factors to make $a=1$.
2. Factor the quadratic.
3. Solve for $x$ by setting each factor equal to $0$.

Let's work through the first example from the video.

$$x^2-9x+14=0$$

Steps 1 and 1.1 are already done for us! The equation is set equal to $0$ and $a=1$. Let's move on to factoring the equadratic.

$$\begin{align} x^2-9x+14&=0\\ (x-2)(x-7)&=0\end{align}$$

The last step is to set each factor equal to $0$ and solve for $x$.

$$\begin{align}(x-2)(x-7)&=0\\ \\ x-2&=0\\ x&=2\\ \\ x-7&=0\\ x&=7\end{align}$$

The solutions are $x=2,7$.

The Fraction Method for Factoring

We can use the Fraction Method, based on Vieta's Theorem, to factor more efficiently than using guess-and-check.

1. Move everything to one side and identify your coefficients ===

For example, if you were given: $$x^{2}+2x=-1$$ You would want to add $1$ to both sides in order to make the equation look like this: $$x^{2}+2x+1=0$$ And then identify your coefficients: $$\begin{align}a&=1\\ b&=2\\ c&=1\end{align}$$ Your next step is:

1. Multiply the $a$ value and the $c$ value together ===

So, for the equation we used in the last step, we would do this: $$a \times c=(1) \times (1)=1$$

We'll call this $m$. You want to to find two values that multiply to equal $m$ and add to equal $b$. So your third step is:

1. Find two values that multiply to get $m$ and add them to equal $b$ ===

We need to find two values that multiply to get $1$ and add to get $2$, so:

$$\begin{align} j+k&=2\\ &\text{and}\\ j \times k&=1\end{align}$$

With this, we can see that $j$ an d $k$ both must be $1$ since $1+1=2$ and $1 \times 1=1$. Now that we have these two values, we will "divide" both of them by $a$.

1. Divide $j$ and $k$ by $a$ ===

$$\frac{j}{a} \quad \quad \quad \frac{k}{a}$$ $$\Downarrow \quad \quad \quad \Downarrow$$ $$\frac{1}{1} \quad \quad \quad \frac{1}{1}$$

The next step is very important to correctly factoring this equation.

1. Let the denominator of each fraction be the coefficient of the $x$ value. Let the numerator be what you add to that $x$ value ===

For the first fraction, the denominator$=1$ and the numerator$=1$ so, $$\frac{1}{1} \Longrightarrow (1x+1)$$ For the second fraction, the denominator$=1$ and the numerator$=1$ so, $$\frac{1}{1} \Longrightarrow (1x+1)$$ You have both components of the equation in factored form. Next,

1. Multiply the two components together and set it equal to $0$ ===

$$(1x+1) \times (1x+1)=0$$

Below are examples of different problems to help you understand them more.

Example 1

Example 2

Example 3

Solving Quadratic Equations by Factoring

There are many ways to solve a quadratic equation, including graphing, using the quadratic formula, and factoring. Factoring is the easiest and quickest way to solve a quadratic function, but it doesn't work for everything. Let's look at a very simple example of factoring a quadratic in standard form $y=ax^2+bx+c$. The steps boil down to:

1. Make sure one side of the solution is equal to zero. You may have to manipulate the equation into standard form.

2. If possible, factor out constants to get $a=1$. These next steps are only for if the leading coefficient, $a$, $a=1$. $a$ can't equal a negative number.

3. Find two numbers that add up to $b$ and multiply to $c$ ($c$ is the constant). These then go into the form: $0=(x+\;\;)(x+\;\;)$. We do this any time we factor a polynomial.

4. Last, we set each of the two terms in parentheses equal to $0$ and solve for $x$. The Zero Product Property allows to solve for $x$ with both factors.

Let's take a look at an example.

Solve $4x^2-15x=-16$

We can start by factoring out the common factor, $4$.

$$\begin{align}\frac{4x^2-15x}{4}&=\frac{-16}{4}\\ x^2-5x&=-4\end{align}$$

Now we're going to make sure one side is $0$.

$$\begin{align} x^2-5x&=-4\\ x^2-5x+4&=0\end{align}$$

Let's factor the left side.

$$\begin{align}x^2-5x+4&=0\\ (x-4)(x-1)&=0\end{align}$$

We're going to use the zero product property to solve each factor for $x$.

$$\begin{align}(x-4)(x-1)&=0\\ \\ x-4&=0\\ x&=4\\ \\ x-1&=0\\ x&=1\end{align}$$

Our solutions are $x=4,1$.