Quadratic Equation Factorization

Finding Roots from Linear Factors

Factoring rewrites a quadratic equation from $ax^2 + bx + c = 0$ as a product. When it can be written as $a(x - p)(x - q) = 0$, the numbers $p$ and $q$ are its roots.

The roots are the values of $x$ that make the equation equal to zero. In product form, we can find them directly by setting each factor equal to zero.

Basic Principles of Factorization

A quadratic equation in standard form is written as:

In this standard form, $a$, $b$, and $c$ are constants and $a \neq 0$.

Factorization uses the zero-product property: if a product equals zero, at least one factor must equal zero.

This means if $(x - p)(x - q) = 0$, then:

• $x - p = 0$ or $x - q = 0$
• Therefore $x = p$ or $x = q$

Steps for Factoring Quadratic Equations

For a quadratic equation $ax^2 + bx + c = 0$ with integer coefficients, factoring by grouping follows these steps:

1. Ensure the equation is in standard form with the right side equal to zero
2. Find two numbers that when multiplied give $ac$ and when added give $b$
3. Split the middle term with those two numbers, then factor by grouping
4. Determine the roots of the equation from these factors

Examples of factoring quadratic equations

The three examples below climb the same four steps: first with $a = 1$, then with a negative constant term, then with $a \neq 1$. Watch for the factor pair with product $ac$ and sum $b$ in each one.

1. Factoring the equation:

   $$x^2 + 5x + 6 = 0$$

   In this equation, $a = 1$, $b = 5$, and $c = 6$.

   Step 1: The equation is already in standard form with the right side equal to zero.

   Step 2: We need to find two numbers that:

   • When multiplied give $ac = 1 \times 6 = 6$
   • When added give $b = 5$

   Factors of $6$ are $1$, $2$, $3$, and $6$. Possible factor pairs are $(1, 6)$ and $(2, 3)$. The pair $(2, 3)$ gives a sum of $5$, which matches the value of $b$.

   Step 3: We can write the equation as:

   $$x^2 + 2x + 3x + 6 = 0$$

   $$x(x + 2) + 3(x + 2) = 0$$

   $$(x + 3)(x + 2) = 0$$

   Step 4: From the factored form above, we get:

   • $x + 3 = 0$ → $x = -3$
   • $x + 2 = 0$ → $x = -2$

   Therefore, the roots of the equation are $x = -3$ and $x = -2$.

2. Factoring the equation:

   $$3x^2 + 13x - 10 = 0$$

   In this equation, $a = 3$, $b = 13$, and $c = -10$.

   Step 1: The equation is already in standard form with the right side equal to zero.

   Step 2: We need to find two numbers that:

   • When multiplied give $ac = 3 \times (-10) = -30$
   • When added give $b = 13$

   Factors of $-30$ are pairs of numbers with opposite signs:

   $$\begin{aligned} &(1, -30), \; (-1, 30), \; (2, -15), \\ &\; (-2, 15), \\ &(3, -10), \; (-3, 10), \; (5, -6), \\ &\; (-5, 6) \end{aligned}$$

   The pair $(15, -2)$ gives a sum of $13$, which matches the value of $b$.

   Step 3: We can write the equation as:

   $$3x^2 + 15x - 2x - 10 = 0$$

   We can group the terms:

   $$3x^2 + 15x - 2x - 10 = 0$$

   $$3x(x + 5) - 2(x + 5) = 0$$

   $$(3x - 2)(x + 5) = 0$$

   Step 4: From the factored form above, we get:

   • $3x - 2 = 0$ → $x = \frac{2}{3}$
   • $x + 5 = 0$ → $x = -5$

   Therefore, the roots of the equation are $x = \frac{2}{3}$ and $x = -5$.

3. Factorization When Coefficient $a \neq 1$

   When the coefficient $a$ is not equal to $1$, we need some modifications in the factorization steps. There are several approaches:

   Method Using Factors of $ac$

   1. Determine the value of $ac$
   2. Find a pair of factors of $ac$ that when added give $b$
   3. Use this factor pair to split the term $bx$ into two terms
   4. Factor by grouping

   Example of Factorization:

   $$2x^2 + 5x - 3 = 0$$

   In this equation, $a = 2$, $b = 5$, and $c = -3$.

   Step 1: Calculate $ac = 2 \times (-3) = -6$

   Step 2: Find a pair of factors of $-6$ that when added give $5$:

   Factors of $-6$: $(1, -6), (-1, 6), (2, -3), (-2, 3)$

   The pair $(6, -1)$ gives a sum of $5$, which matches the value of $b$.

   Step 3: Split the term $5x$ into $6x - x$:

   $$2x^2 + 6x - x - 3 = 0$$

   Step 4: Factor by grouping:

   $$2x^2 + 6x - x - 3 = 0$$

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

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

   Step 5: Determine the roots of the equation:

   • $2x - 1 = 0$ → $x = \frac{1}{2}$
   • $x + 3 = 0$ → $x = -3$

The roots of the equation are $x = \frac{1}{2}$ and $x = -3$.

Factoring When One Root Is Known

If we know one of the roots of a quadratic equation, we can use this information to find the complete factorization.

Example: One of the roots of the equation $2x^2 - bx - 6 = 0$ is $6$

If $x = 6$ is a root of the equation, then $(x - 6)$ is one of its factors.

We can substitute $x = 6$ into the original equation:

Now we can write the equation as $2x^2 - 11x - 6 = 0$.

Using the factorization method, we factor it as:

The roots of the equation are $x = -\frac{1}{2}$ and $x = 6$.

Special Cases of Factorization

1. Form $ax^2 + bx = 0$

   For equations without a constant term, we can factor out $x$ directly:

   $$ax^2 + bx = 0$$

   $$x(ax + b) = 0$$

   The roots are $x = 0$ and $x = -\frac{b}{a}$.

   Example: $2x^2 - 18x = 0$

   $$2x^2 - 18x = 0$$

   $$2x(x - 9) = 0$$

   The roots are $x = 0$ and $x = 9$.

2. Form $ax^2 - c = 0$

   For equations without an $x$ term, first isolate $x^2$. Over the real numbers, the square-root step requires $\frac{c}{a}\ge0$:

   $$ax^2 - c = 0$$

   $$ax^2 = c$$

   $$x^2 = \frac{c}{a}$$

   $$x = \pm \sqrt{\frac{c}{a}}$$
   • If $\frac{c}{a}>0$, there are two distinct real roots and the expression factors into real linear factors.
   • If $\frac{c}{a}=0$, $x=0$ is a repeated root.
   • If $\frac{c}{a}<0$, there are no real roots and no factorization into real linear factors. Over the complex numbers, the roots are $x=\pm i\sqrt{-\frac{c}{a}}$.

   Example: $2x^2 - 18 = 0$

   $$2x^2 - 18 = 0$$

   $$2x^2 = 18$$

   $$x^2 = 9$$

   $$x = \pm 3$$

   The roots are $x = 3$ and $x = -3$.

   Example with complex roots: $2x^2 + 18 = 0$

   $$2x^2 + 18 = 0$$

   $$2x^2 = -18$$

   $$x^2 = -9$$

   The roots are $x = 3i$ and $x = -3i$. Check against the formula above: $c/a = -9$, so $x = \pm i\sqrt{9} = \pm 3i$.

When Integer Factor Pairs Do Not Work

Not every quadratic equation has rational factors. When no suitable factor pair exists, use another method such as the quadratic formula:

For integer coefficients, the quadratic has rational roots exactly when the discriminant $b^2 - 4ac$ is a nonnegative perfect square.

Practice Problems

Each equation has to reach factored form first, so look for two factors whose product gives the constant term. The solutions then follow directly from those two factors.

Factor the following quadratic equations:

1. $x^2 + 6x + 5 = 0$
2. $2x^2 + 5x + 2 = 0$
3. $3x^2 - x - 2 = 0$
4. $2x^2 - 8x + 6 = 0$
5. $x^2 - 9 = 0$

Worked Solutions

1. $x^2 + 6x + 5 = 0$

   Step 1: Identify the coefficients

   $$a = 1, b = 6, c = 5$$

   Step 2: Find two numbers that when multiplied give $ac = 5$ and when added give $b = 6$

   Choose factors of $5$ with the required sum:

   $$\begin{aligned} &1\times5=5 \\ &1+5=6 \end{aligned}$$

   Step 3: Factorization

   $$x^2 + 6x + 5 = 0$$

   $$x^2 + x + 5x + 5 = 0$$

   $$x(x + 1) + 5(x + 1) = 0$$

   $$(x + 5)(x + 1) = 0$$

   Step 4: Determine the roots of the equation

   $$x + 5 = 0 \Rightarrow x = -5$$

   $$x + 1 = 0 \Rightarrow x = -1$$

   The roots of the equation are $x = -5$ and $x = -1$.

2. $2x^2 + 5x + 2 = 0$

   Step 1: Identify the coefficients

   $$a = 2, b = 5, c = 2$$

   Step 2: Find two numbers that when multiplied give $ac = 2 \times 2 = 4$ and when added give $b = 5$

   Choose factors of $4$ with the required sum:

   $$\begin{aligned} &1\times4=4 \\ &1+4=5 \end{aligned}$$

   Step 3: Factorization

   $$2x^2 + 5x + 2 = 0$$

   $$2x^2 + x + 4x + 2 = 0$$

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

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

   Step 4: Determine the roots of the equation

   $$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$$

   $$x + 2 = 0 \Rightarrow x = -2$$

   The roots of the equation are $x = -\frac{1}{2}$ and $x = -2$.

3. $3x^2 - x - 2 = 0$

   Step 1: Identify the coefficients

   $$a = 3, b = -1, c = -2$$

   Step 2: Find two numbers that when multiplied give $ac = 3 \times (-2) = -6$ and when added give $b = -1$

   Choose factors of $-6$ with the required sum:

   $$\begin{aligned} &(-3)\times2=-6 \\ &(-3)+2=-1 \end{aligned}$$

   Step 3: Factorization

   $$3x^2 - x - 2 = 0$$

   $$3x^2 - 3x + 2x - 2 = 0$$

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

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

   Step 4: Determine the roots of the equation

   $$3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$$

   $$x - 1 = 0 \Rightarrow x = 1$$

   The roots of the equation are $x = -\frac{2}{3}$ and $x = 1$.

4. $2x^2 - 8x + 6 = 0$

   Step 1: Identify the coefficients

   $$a = 2, b = -8, c = 6$$

   Step 2: Find two numbers that when multiplied give $ac = 2 \times 6 = 12$ and when added give $b = -8$

   Choose factors of $12$ with the required sum:

   $$\begin{aligned} &(-6)\times(-2)=12 \\ &(-6)+(-2)=-8 \end{aligned}$$

   Step 3: Factorization

   $$2x^2 - 8x + 6 = 0$$

   $$2x^2 - 6x - 2x + 6 = 0$$

   $$2x(x - 3) - 2(x - 3) = 0$$

   $$(2x - 2)(x - 3) = 0$$

   $$2(x - 1)(x - 3) = 0$$

   Step 4: Determine the roots of the equation

   $$x - 1 = 0 \Rightarrow x = 1$$

   $$x - 3 = 0 \Rightarrow x = 3$$

   The roots of the equation are $x = 1$ and $x = 3$.

5. $x^2 - 9 = 0$

   Step 1: Identify as a difference of squares

   $$x^2 - 9 = x^2 - 3^2$$

   Step 2: Use the difference of squares formula $a^2 - b^2 = (a+b)(a-b)$

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

   Step 3: Determine the roots of the equation

   $$x + 3 = 0 \Rightarrow x = -3$$

   $$x - 3 = 0 \Rightarrow x = 3$$

   The roots of the equation are $x = -3$ and $x = 3$.