Hi. This page will be about factoring quadratic equations when a = 1 in \(y = ax^2 + bx + c\).

 

Table Of Contents

 

 

Introduction

 

Another factoring method for quadratic equations will be shown here. We deal with the case of \(a = 1\) in \(y = ax^2 + bx + c\). If \(a \neq 1\) then another method may be needed. (Also if \(a = 0\) then we no longer have a quadratic equation.)

It is assumed that the reader is familiar with common factoring.

 

The Factoring Method

 

To illustrate the method, an example will be used.

Suppose that we are given \(y = x^2 + 5x + 4\). We need two numbers \(j\) and \(k\) which are factors of 4 and satisfy \(j \times k\) = 4 and \(j + k = 5\). The two numbers which fit that criteria are 1 and 4 (or 4 and 1) since \(1 \times 4 = 4\) and \(4 + 1 = 5\).

 

Using the numbers 1 and 4 we can factor \(y = x^2 + 5x + 4\) into \(y = (x + 1)(x + 4)\). The equation \(y = (x + 1)(x + 4)\) follows the factored form format of \(y = (x + j)(x + k)\).

 

To check that \(y = (x + 1)(x + 4)\) is indeed the factored form of \(y = x^2 + 5x + 4\), we use the FOIL method when multiplying binomials.

 

\[y = (x + 1)(x + 4) = x^2 + 4x + x + 4 = x^2 + 5x + 4\]

 

There are times when \(j = k\). As an example, the factored form of \(y = x^2 +2x + 1\) is \(y = (x + 1)(x + 1) = (x + 1)^2\).

 

The General Method

Given a quadratic equation of the form \(y = ax^2 + bx + c\), we can factor it into the form \(y = (x + j)(x + k)\). To determine what \(j\) and \(k\), we seek two numbers \(j\) and \(k\) such that \(jk = ac = 1c = c\) (since \(a =1\) here and \(j + k = b\)).

The examples below will illustrate how the general method works.

 

Examples

 

Example One

Factor \(y = x^2 + 7x + 12\).

Answer:

Two factors of 12 which multiply to 12 are 1 and 12, 2 and 6, 3 and 4 (you can include the reverse ordered pairs too). Out of the pairs which multiply to 12, 3 and 4 sum to 7. The quadratic equation would be factored as \(y = (x + 3)(x + 4)\).

Checking Our Answer:

 

\[y = (x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12\]

 


Example Two

Factor \(y = x^2 + 20x + 99\).

 

Factors of 99 are 1 and 99, 3 and 33, 9 and 11. The two numbers which multiply together to get 99 and add together to get 20 are 9 and 11. The factored form of \(y = x^2 + 20x + 99\) would be \(y = (x + 9)(x + 11)\).

Checking Our Answer:

 

\[y = (x + 9)(x + 11) = x^2 + 11x + 9x + 99 = x^2 + 20x + 99\]

 


Example Three

Factor \(y = -x^2 - 3x - 2\) by factoring out \((-1)\) as a common factor first.

Answer:

 

\[y = -x^2 - 3x - 2 = -1(x^2 + 3x + 2)\]

 

From factoring out the (-1), we have a factorable quadratic in the brackets. The only factors of 2 are 1 and 2. We can factor \(y = -1(x^2 + 3x + 2)\) into \(y = -(x + 2)(x + 1)\).

Checking Our Answer:

 

\[y = -(x + 2)(x + 1) = -(x^2 + x + 2x + 2) \]

\[y = -(x^2 + 3x + 2) \]

\[ y = -x^2 -3x -2 \]

 


Example Four

Factor \(y = x^2 - x - 6\).

Answer:

In this equation we have \(a = 1\), \(b = -1\) and \(c = -6\) in \(y = ax^2 + bx + c\). Unlike examples one and two, we have to be careful with signs. I prefer to ignore the negative and look at the factors of \(c = 6\).

Factors of 6 are 1 and 6, 2 and 3. Now, we consider the signs. One of the two numbers in the factor pair is negative. The pair 6 and 1 would not work as 6 - 1 = 5 and 1 - 6 = -5. With 2 and 3, we have 2 - 3 = -1 and 3 - 2 = 1. We go with 2 and -3.

The factored form of \(y = x^2 - x - 6\) is \(y = (x + 2)(x - 3)\).

Checking Our Answer:

 

\[y = (x + 2)(x - 3) = x^2 - 3x + 2x + (2)(-3) = x^2 - x - 6\]

 

Practice Problems

 

Here are some practice problems to build understanding.

  1. Factor \(y = x^2 + 4x + 4\).

  2. Factor \(y = x^2 + 5x + 6\).

  3. Factor \(y = x^2 + 25x + 100\).

  4. Factor \(y = x^2 - x - 12\).

  5. Factor \(y = x^2 - 9x + 20\).

  6. In \(y = -x^2 - 6x - 5\), factor a (-1) first and then factor using the method described in this page.

  7. Using common factoring first, factor \(y = 3x^2 + 15x + 18\).

 

Answers

 

  1. \(y = x^2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)^2\)

  2. \(y = x^2 + 5x + 6 = (x + 3)(x + 2)\)

  3. \(y = x^2 + 25x + 100 = (x + 20)(x + 5)\)

  4. \(y = x^2 - x - 12 = (x - 4)(x + 3)\)

  5. \(y = x^2 - 9x + 20 = (x - 5)(x - 4)\)

  6. \(y = -x^2 - 6x - 5 = -1(x^2 + 6x + 5) = -(x + 5)(x + 1)\)

  7. \(y = 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 3)(x + 2)\)