Hi. This page will be about factoring quadratic equations when a = 1 in \(y = ax^2 + bx + c\).
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.
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.
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:
\[y = (x + 2)(x - 3) = x^2 - 3x + 2x + (2)(-3) = x^2 - x - 6\]
Here are some practice problems to build understanding.
Factor \(y = x^2 + 4x + 4\).
Factor \(y = x^2 + 5x + 6\).
Factor \(y = x^2 + 25x + 100\).
Factor \(y = x^2 - x - 12\).
Factor \(y = x^2 - 9x + 20\).
In \(y = -x^2 - 6x - 5\), factor a (-1) first and then factor using the method described in this page.
Using common factoring first, factor \(y = 3x^2 + 15x + 18\).
\(y = x^2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)^2\)
\(y = x^2 + 5x + 6 = (x + 3)(x + 2)\)
\(y = x^2 + 25x + 100 = (x + 20)(x + 5)\)
\(y = x^2 - x - 12 = (x - 4)(x + 3)\)
\(y = x^2 - 9x + 20 = (x - 5)(x - 4)\)
\(y = -x^2 - 6x - 5 = -1(x^2 + 6x + 5) = -(x + 5)(x + 1)\)
\(y = 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 3)(x + 2)\)