Tutorial: Factoring quadratics

Tutorial: Using exponent identities

Adaptive game version

Go to Part A: Multiplying algebraic expressions

Go to Part B: Factoring algebraic expressions: Taking out common factors

This tutorial: Part C: Factoring quadratics

(This topic is also in Section 0.3 in Finite Mathematics and Applied Calculus)

What is an adaptive game tutorial? ▶

In Part A of this tutorial we saw how to use the distributive law or FOIL to do calculations like

$(x + 2)(2x - 5) = 2x^2 - x - 10.$ In this section, we want to reverse the process; that is, starting with the expression $2x^2 - x - 10,$ we would like to factor it and get back the original expression $(x + 2)(2x - 5).$

There are some quadratic expressions, such as $x^2 + x + 1,$ that cannot be factored in this form at all. Here, we shall consider only quadratic expressions that do factor, and in such a way that the numbers $d, e, f$ and $g$ are integers (whole numbers). Other cases are fully discussed in the %%solvingPolytut.

Factoring by trial and error

The usual technique of factoring such quadratic expressions is a "trial-and-error" approach, which we illustrate by means of an example and some exercises for you.

Factoring by trial and error: %%Example

#[Let us factor][Factorizemos]# $x^2 - 6x + 5.$

Solution
Find ways to factor the first and last terms:

First term: \t $x^2$ has factors $\color{#0ea05e}{x}$ %%and $\color{#de6c00}{x}$ \t because $\color{slateblue}{x \cdot x = x^2}$ \\ Last term: \t $5$ has factors $\color{#c1026f}{5}$ %%and $\color{#026fc1}{1}$ \t because $\color{slateblue}{5 \cdot 1 = 5}$

#[Group them together and make an attempt.][Agrúpalos juntos y haz un intento]#:

$(\color{#0ea05e}{x} + \color{#c1026f}{5})(\color{#de6c00}{x} + \color{#026fc1}{1}) = x^2 + 6x + 5$

This is fine, except for the sign of the middle term. But notice that we can also get the $5$ by multiplying $\color{#c1026f}{(-5)}$ and $\color{#026fc1}{(-1)}.$ In other words, $5$ also has factors $\color{#c1026f}{(-5)}$ and $\color{#026fc1}{(-1)}.$ Using these instead gives

$(\color{#0ea05e}{x} \color{#c1026f}{- 5})(\color{#de6c00}{x} \color{#026fc1}{- 1}) = x^2 - 6x + 5,$

so we have found the correct factorization.

Suggested video for this topic: Video by ThinkwellVids

#[Solving quadratic equations by factoring][Resolver ecuaciones cuadráticas por factorizar]#

%%Q #[What is the point of all of this factoring anyway?][ ¿Cuál es el punto de toda esta factorización de todos modos?]#
%%A #[The most common application is to use it to solve equations of the form][La aplicación más común es utilizarla para resolver ecuaciones de la forma]#

#[Quadratic equation][Ecuación cuadrática]#

#[Before we start solving quadratic equations, let's warm up by first solving linear equations:][Antes de empezar a resolver ecuaciones cuadráticas, primero vamos a calentar mediante la resolución de ecuaciones lineales:]#

#[Warmup: Solving linear equations][Calentamiento: Resolver ecuaciones lineales]#
#[An important step in solving a quadratic equation is to know how to solve so-called linear equations. A linear equation is an equation of the form][Un paso importante en la solución de una ecuación cuadrática es saber cómo resolver las llamadas ecuaciones lineales. Una ecuación lineal es una ecuación de la forma]#

$ax + b = c, \qquad $ ($a, b, c$ #[constants with $a$ non-zero.][constantes con $a$ distinta de cero]#)

%%Examples

$3x + 2 = 0 \qquad$ \t $\color{slateblue}{(a = 3,\ b = 2,\ c = 0)}$ \\ $x + 1 = -4$ \t $\color{slateblue}{(a = 1,\ b = 1,\ c = -4)}$ \\ $-x - 6 = 1$ \t $\color{slateblue}{(a = -1,\ b = -6,\ c = 1)}$ \\ $8x = 0$ \t $\color{slateblue}{(a = 8,\ b = 0,\ c = 0)}$

\t !3!#[Solution of][Solución de]# ax + b = c \t !3! #[Eg.][Ej.]# −2x + 5 = 4 \\ \t !5! 1. #[Subtract $b$ from both sides][Restar $b$ de ambos lados]#: (#[If b is negative, this amounts to adding a number to both sides.][Si b es negativo, eso equivale a sumar un número a ambos lados.]#) \\ \\ \t \gap[10] \t !r! $ax + b \color{red}{\ - \ b}$ \t $= c \color{red}{\ - \ b} \qquad$ \t !r! $\color{slateblue}{-2x+5} \color{red}{\ - \ 5} $ \t $\color{slateblue}{= 4} \color{red}{\ - \ 5}$ \\ \t \t !r! $ax $ \t $= c - b$ \t !r! $\color{slateblue}{-2x}$ \t $\color{slateblue}{= -1}$ \t \t \gap[30] \t \\ \t \\ \t !5! 2. #[Divide both sides by $a$][Dividir ambos lados por $a$]#: \\ \\ \t \t !r! $\dfrac{ax}{\color{red}{a}}$ \t $= \dfrac{c-b}{\color{red}{a}} \qquad$ \t !r! $\dfrac{-2x}{\color{red}{-2}}$ \t $=\dfrac{-1}{\color{red}{-2}}$ \\ \t \t !r! $x$ \t $= \dfrac{c-b}{\color{red}{a}} \qquad \qquad$ \t !r! $x$ \t $=\dfrac{1}{2}$

Suggested video for this topic: Video by patrickJMT

#[Some for you to do][Algunas a probar para ti]#

Now we are ready to start solving quadratic equations (We will study solving more general equations starting with %%solvingPolytut.)

#[Solving quadratic equations][Resolver ecuaciones cuadráticas]#: %%Example
Let us solve $2x^2 - 9x + 4 = 0.$

Solution
First, factor the expression using the above techniques:

$2x^2 - 9x + 4 = (2x-1)(x-4)$

So now we can rewrite our equation as

$(2x-1)(x-4) = 0.$

Thus, the product of the two quantities $(2x-1)$ and $(x-4)$ is zero. Now, if a product of two numbers is zero, it means that one or the other of them must be zero. In other words,

Either \gap[5] \t $2x - 1 = 0,$ \gap[5] \t so $2x = 1,$ giving $x = \frac{1}{2},$ \gap[40] \t See the Warmup above. \\ %%or \t $x-4 = 0,$ \t giving $x = 4.$

Thus, the quadratic equation $2x^2 - 9x + 4 = 0$ has two solutions: $x = \frac{1}{2}, \ \ x = 4.$

Suggested video for this topic: Video by patrickJMT

Some for you to do

Now try the exercises in Section 0.3 in Finite Mathematics and Applied Calculus. or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.