# Factoring a Quadratic Polynomial, Part 2

In this section, we're going to look at the "general" quadratic equation. In our first discussion of this kind of polynomial, we looked only at the case where the first or "leading" coefficient was one. That's a special case that's relatively easy to handle. In this lesson, we'll look at the more general case we're the leading coefficient can be any number.

There are two basic approaches to this sort of problem. The first is similar to the method we used in earlier discussion: You look at all the factors of the first coefficient and all the factors of the last coefficient and try to find a pair that, when you multiply them together and then add the results, you get the middle term. There are often a lot of possibilities that you have to consider with that approach so it quickly becomes very messy. The approach we're going to discuss and suggest that you use, is much more straight forward and relies on the "factoring by grouping" method that we discussed in the last lesson. The procedure works like this:

Factor: |
6x^{2} + 19x + 10 |

1. Multiply the leading coefficient and the constant at the end. |
6 · 10 = 60 |

2. Find two factors of the number you found in step one that add up to the coefficient of the . In our example, we need two numbers that multiply to 60 and add up to 19.x |
15 · 4 = 60 15 + 4 = 19 |

3. Split the x term into two parts using the two numbers you found in step three. |
6x^{2} + 15x + 4x + 10 |

4. Factor by grouping. |
3x(x + 5) + 2(2x + 5)(3 x + 2)(2x + 5) |

# Example 1

**Factor 2 x^{2} + 21x + 10.**

We need to start by multiplying the first and last numbers in the polynomial together. That means 2 · 10 = 20. Now we need to find two numbers that multiply to give 20 but add up to 21 (the coefficient of the x term). A little trial and error shows us that we're looking for 20 and 1 since 20 · 1 = 20 and 20 + 1 = 21.

The next step is to rewrite the middle term of the original polynomial, 21*x*, in terms of 20 and 1. That would mean

21*x* = 20*x* + 1*x*

Doing that in the original polynomial gives us

2*x*^{2} + 21*x* + 10 = 2*x*^{2} + 20*x* + 1*x* + 10

The expression on the right hand side is one we can factor by grouping to get our final answer.

2*x*^{2} + 20*x* + 1*x* + 10

2*x*(*x* + 10) + (*x* + 10)

(2*x* + 1)(*x* + 10)

# Example 3

**Factor 15 x^{2} - 26x + 8.**

In this example, we start off with 15 · 8 = 120 which has a lot of factors. With a number as big as this, it may be helpful for you to make a complete list like we did in Example 2. If you do, be sure that you consider the negative possibilities as well as the positive ones, e.g. -60 · -2. If you make the list and sort thorugh it you'll find that -20 · -6 = 120 and -20 + -6 = -26 which is the coefficient of the *x* term in our example.

Splitting the -26*x* into -20*x* and -6*x* gives us

15*x*^{2} - 20*x* - 6*x* + 8

Which we can factor to

5*x*(3*x* - 4) - 2(3*x* - 4)

(3*x* - 4)(5*x* - 2)

# Example 2

**Factor 4 x^{2} + 17x - 15.**

This example follows the same pattern as the Example 1. First we need to multiply together the first and last coefficients: 4 · -15 = -60. Now we need two numbers that multiply to equal -60 but that add up to 17. It might be obvious to you that +20 and -3 do that. If it isn't then write out all the numbers that multiply out to -60 like we did in the "Factoring Trinomials 1" lesson and add them all up:

1 x -60 | 2 x -30 | 3 x -20 |

4 x -15 | 5 x -12 | 6 x -10 |

-1 x 60 | -2 x 30 | -3 x 20 |

-4 x 15 | -5 x 12 | -6 x 10 |

Now we'll find the sums of each pair.

1 - 60 = -59 | 2 - 30 = -28 | 3 - 20 = -17 |

4 - 15 = -11 | 5 - 12 = -7 | 6 - 10 = -4 |

-1 + 60 = 59 | -2 + 30 = 28 | -3 + 20 = 17 |

-4 + 15 = 11 | -5 + 12 = 7 | -6 + 10 = 4 |

Look over the bottom list and you'll see that 20 and -3 are the only pair that add up to 17 so those are the ones that we need to rewrite 17*x* in terms of:

17*x* = 20*x* - 3*x*

Doing that in the original polynomial gives us something we can factor by grouping.

4*x*^{2} + 20*x* - 3*x* - 15

4*x*(*x* + 5) - 3(*x* + 5)

(4*x* - 3)(*x* + 5)

# Example 4

**Factor -2 x^{2} + x + 28.**

This one may look a little confusing because of the negative sign in front of the *x*^{2} term but it works the same as all of the others. First we look at -2 · 28 = -56 and then we notice that 8 · -7 = -56 and 8 + -7 = 1. (Remember that if there's no number written out in front of a variable then its coefficient is 1.)

Now we can rewrite *x* as 8*x* - 7*x* to get

-2*x*^{2} + 8*x* - 7*x* + 28

-2*x*(*x* - 4) - 7(*x* - 4)

(*x* - 4)(-2*x* - 7)

**Directions:** This solution has 5 steps. To see a description of each step click on the boxes on the left side below. To see the calculations, click on the corresponding box on the right side. Try working out the solution yourself and use the descriptions if you need a hint and the calculations to check your solution.