Factoring Quadratic Equations (a = 1 Case)

Topic: Factoring Quadratic Equations When a = 1

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 inculde 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:

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


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)

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

7)

    \[y = 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 3)(x + 2)\]

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