- Intro to grouping
- Factoring by grouping
- Factoring quadratics by grouping
- Factoring quadratics: leading coefficient ≠ 1
- Factor quadratics by grouping
- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Creativity break: How can we combine ways of thinking in problem solving?
Learn how to factor quadratic expressions as the product of two linear binomials. For example, 2x²+7x+3=(2x+1)(x+3).
What you need to know before taking this lesson
The grouping method can be used to factor polynomials with terms by taking out common factors multiple times. If this is new to you, you'll want to check out our Intro to factoring by grouping article.
We also recommend that you review our article on factoring quadratics with a leading coefficient of 1 before proceeding.
What you will learn in this lesson
In this article, we will use grouping to factor quadratics with a leading coefficient other than , like .
Example 1: Factoring
Since the leading coefficient of is , we cannot use the sum-product method to factor the quadratic expression.
Instead, to factor , we need to find two integers with a product of (the leading coefficient times the constant term) and a sum of (the -coefficient).
Since and , the two numbers are and .
These two numbers tell us how to break up the -term in the original expression. So we can express our polynomial as .
We can now use grouping to factor the polynomial:
The factored form is .
In general, we can use the following steps to factor a quadratic of the form :
- Start by finding two numbers that multiply to and add to .
- Use these numbers to split up the -term.
- Use grouping to factor the quadratic expression.
Check your understanding
1) Factor .
2) Factor .
Example 2: Factoring
To factor , we need to find two integers with a product of and a sum of .
Since and , the numbers are and .
We can now write the term as the sum of and and use grouping to factor the polynomial:
The factored form is .
Take note: In step above, notice that because the third term is negative, a "+" was inserted between the groupings to keep the expression equivalent to the original. Also, in step , we needed to factor out a negative GCF from the second grouping to reveal a common factor of . Be careful with your signs!
Check your understanding
3) Factor .
4) Factor .
5) Factor .
When is this method useful?
Well, clearly, the method is useful to factor quadratics of the form , even when .
However, it's not always possible to factor a quadratic expression of this form using our method.
For example, let's take the expression . To factor it, we need to find two integers with a product of and a sum of . Try as you might, you will not find two such integers.
Therefore, our method doesn't work for , and for a bunch of other quadratic expressions.
It's useful to remember, however, that if this method doesn't work, it means the expression cannot be factored as where , , , and are integers.
Why is this method working?
Let's take a deep dive into why this method is at all successful. We will have to use a bunch of letters here, but please bear with us!
Suppose the general quadratic expression can be factored as with integers , , , and .
When we expand the parentheses, we obtain the quadratic expression .
Since this expression is equivalent to , the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:
Now, let's define and .
According to this definition...
And so and are the two integers we are always looking for when we use this factorization method!
The next step in the method after finding and is to split the -coefficient according to and and factor using grouping.
Indeed, if we split the -term into , we will be able to use grouping to factor our expression back into .
In conclusion, in this section we...
- started with the general expanded expression and its general factorization ,
- were able to find two numbers, and , such that and we did so by defining and ,
- split the -term into , and were able to factor the expanded expression back into .
This process shows why, if an expression can indeed be factored as , our method will ensure that we find this factorization.
Thanks for pulling through!
Want to join the conversation?
- So, this metod is not applicable to this equation: 3x^2+5x+10 for example. What method can I use to factorize this equation? Or there are any quadratics can´t be factorizated?(8 votes)
- The quadratic formula always works, so first look at the discriminant b^2 - 4 a c
For your equation 5*2 - 4(3)(10) or 25 - 120 = - 95
Since you cannot take SQRT of negative number in real domain
This cannot be factorable since there is no real solution(55 votes)
- So on number 5, How do you know to group -4 with 6x^ or -9 with it?(4 votes)
- On question 4)factor 3x^2 -2x- 5
In working it out I got the answer (x-1)(3x+5) where as the correct answer was (3x-5)(x+1). The difference appears because there were multiple ways to split the GCF. As far as I can tell they both give the right answer and expand back into 3x^2-2x-5. Is this typical when solving these problems or did I do something wrong or is this just a random special case that happens to work?(3 votes)
- Sorry, but your version creates the wrong sign on the middle term. You can see this if you multiply your factors. You get: 3x^2+5x-3x-5. Notice: 5x-3x = 2x, not -2x.
Hope this helps.(10 votes)
- What is the easiest way how to remember how to factor out quadratics?(4 votes)
- There are some songs on youtube to help memorize these formulas. I recommend the quadratic formula set to the tune of "Pop Goes the Weasel", because you can use that formula for a lot of different equations, and the song is very memorable. Hope this answers your question :)(1 vote)
- How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3(1 vote)
- Multiply both sides by 3, which gets -2x-22=(1/2x)-2x-3
Add 2x, so -22=(1/2x)-3. Then add 3, so -19=(1/2x). Multiply by 2x so -38x=1. Then divide by -38, so
x = (-1 / 38)(1 vote)
- How to add squares ?(2 votes)
- To answer this question I will use the example:
4x^2 - 10x + 4
When you take the constant from the squared variable in the front, and you multiply it by the last number, you get 16, and 16 is the result of 4^2, but you don't have to use 4 to get to sixteen. You could also use 8 and 2. -8 - 2 does equal -10, so it works and you get:
4x^2 - 8x - 2x + 4
then you group them
(4x^2 - 8x) + (- 2x + 4)
and factor it
4x(x - 2) + -2(x - 2)
factor it again to get your answer:
(x - 2) (4x - 2)
Hope this answered your question, tell me if it didn't(3 votes)
- With the example -3x^2+17x-20 I found that a=-3 b=20.
This was the working that followed:
However, when I put this solution into the quiz it said I was wrong and in their solution, it substitutes a = 12 and b = 5.
Hence working would be:
Does this have two solutions or have I done something wrong in my working out?(2 votes)
- Several issues, the first is that if you have (20x-20) and factor out a -20, you get -20(-x+1), and thus you do not have the same factor in x+1 and -x+1.
On the correct one, you start with -3x^2+ 12x which factoring out a -3x gives -3x(x-4). So to match, you should have factored out a 5 (not negative 5) to get 5(x-4) so that the x-4 matches to get (-3x+5)(x-4). If one has an x+4 and the other has x-4, you cannot combine them.
So if you expand by FOIL: (-3x+5)(x-4) = -3x^2 + 12x + 5x - 20 = -3x^2 + 17x - 20.
Expanding your first work gives (-3x-20)(x+1)= -3x^2 - 3x - 20x - 20 which gives 3x^2 - 23x - 20 (does not match). Expanding your second gives (-3x+5)(x+4) which gives -3x^2 - 12x + 5x + 20 = -3x^2 - 7x + 20 (does not match). So you struggled both with factoring out numbers in first case and factoring what cannot be factored in the second.(2 votes)
- A continuation of *@Aleman*'s question: Not all expressions can be factorised using this method? If so, how do you know when to use which type?(1 vote)
- There may be a way to tell but I do not know. Honestly I just use the ways that work for all of them. Specifically I use completing the square.(3 votes)