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## Algebra 1

### Course: Algebra 1 > Unit 13

Lesson 9: Strategy in factoring quadratics# Strategy in factoring quadratics (part 2 of 2)

CCSS.Math: , ,

There are a lot of methods to factor quadratics, which apply on different occasions and conditions. After learning all of them in separate, let's think strategically about which method is useful for a given quadratic expression we want to factor.

## Want to join the conversation?

- At7:24, how did he get the +1 to add to the equation when factoring out (x+3)?(8 votes)
- If you have a expression inside of a parentheses without any number in front, it is an invisible 1, so (x+3) is the same as 1(x+3) - distributing a 1 does not change anything. So when Sal factors, he needs the invisible 1 to make sense of the expression.(6 votes)

- I understand how to do this, but I am struggling with this one equation and cannot find anything to help me with it. y=x^2+4x-12(3 votes)
- While factoring quadratics, you have 2 options; grouping and the criss cross method. I personally prefer the criss-cross method for its simplicity. In this question, you'll want to ask yourself: What multiples to -12 but adds up to 4? In this case, you would know that one number would have to be positive while the other is negative, and the positive number will be larger than the absolute negative number. The answer in this question will be -2 and 6 because they fit the requirements.

This is how it would look like in the criss-cross method:

x -2 (Criss cross means the top left multiplying the bottom

x 6 right and adding it up with the top right multiplying with the bottom left)

Therefore, your solution would be (x-2)(x+6). There are methods on Khan Academy and I suggest you learn all of them, especially the criss cross method. I hope this helped!(4 votes)

- question guys;

ex: 1+12x+36x^2

so i did the method of grouping... here are my steps..

36^2+12x+1

(36x^2+6x)(6x+1)

6x(6x+1)(6x+1)

what did i do wrong there..

i know there are other ways of doing this problem but the way i did it, by grouping, where did i go wrong..

i know the answer is (6x+1)^2(2 votes)- The error is in the second step. There should be a plus sign between the parentheses

(36x^2+6x)+(6x+1)

You should have 4 terms (items being added / subtract when you group the pairs.

Then, you find the common factor in each pair

The common factor for: (36x^2+6x) is 6x. Factor it out and you get: 6x(6x+1)+(6x+1)

Then, we need to find the common factor in the 2nd pair. It is 1. Factor it out and you get:

6x(6x+1)+1(6x+1)

We now have 2 terms. The common factor is the binomial (6x+1). Factor it out and you get: (6x+1)(6x+1) or (6x+1)^2

Hope this helps.(7 votes)

- In the form ax^2 + bx + c, where "a" isn't equal to one in the last example, don't you still have to divide by "a" after you're done factoring? Thanks.(1 vote)
- Technically, no. For example you could factor that equation into: (ax+b)(x+c/b)(1 vote)

- What about equations like V^4+v^4x^4?(2 votes)
- Do you mean that you would have an exponent exponented? I don't quite follow you.(2 votes)

- can you solve any quadratic equation with the quadratic formula(2 votes)
- Sort of, sometimes the quadratic formula tells us there are no real solutions, so that would be your answer. we generally do not solve in the imaginary domain,(2 votes)

- Can 7(x^2-9) be 7(x-3)^2?(1 vote)
- No quite because it is the difference of perfect squares, it would be 7(x-3)(x+3).(3 votes)

- How is factoring using for life?(2 votes)
- Lets just pull a zero out of no where... ugh.(1 vote)
- The identity property of addition says that we can add 0 to any number without changing its value. At this point in learning to factor, students are used to seeing quadratics as trinomials (3 terms). So, putting in the missing middle term with a 0x helps students understand how the difference of two squares can be factored if you don't remember the pattern for the special product.(3 votes)

- Something mysterious is going on between4:18and5:35:

a+b and a*b don't seem to be what Sal says they are.

If you write the multiplication of two general binomials you get:

(kx + a)(hx + b) that becomes

hkx^2 + (kb + ah)x + ab.

The formula seems to work pretty well for simple cases where h and k are equal to 1.

But if you apply the formula to this specific scenario things get messy because**a + b**is**not seven**,**ah + bk is seven**!

Following the same reasoning,**ab is not six, abhk is six**.

Can someone help me out?(1 vote)- Sure, so the expression given is 2x^2 + 7x + 3. So your formula was hkx^2 + (kb + ah)x + ab. So this actually still applies. h and k represent factors of the coefficient of x^2, and in this case that coefficient of x^2 is 2. So we know h and k are 2 and 1. The same applies to ab and it is 3, so once again it is the only factors 3 and 1 (it could be negative, but the only way for it to work in this problem was if every h and k were both negative and then it all just cancels out). So we have h and k are 2 and 1, and a and b are 3 and 1. So is there any combination where k*b + a*h equals 7? If we did 3 times 2x it would be 6x and then we would have 1 times 1x which gives us x, we then do 6x + 1x to get 7x, meaning our factors of (2x + 1)(x + 3) works.(3 votes)

## Video transcript

- [Sal] In the last video we looked at three different
examples, really as a bit of a review of some of
our factoring techniques, and also to appreciate when
we might wanna apply them. We saw in the first
example that it was just a process of recognizing a common factor. Once we factored that out, we were done. In the second example, there
was a common factor, four. Then, after that, we used, you could say, our most basic factoring technique, or one of our more basic
factoring techniques, where we say, what two numbers add up to the first-degree coefficient, and then their product is the constant. And we were able to factor the expression. Then, in the third
example, we, once again, started off by factoring
out a common value, which, in this case, was three. We could've done it the same
way we did the second one, or we could've immediately recognized that this is a perfect-square polynomial. But either way, we were able
to factor the expression. Let's keep going to see if we
can tackle some other types of polynomials that might
require some other techniques. Let's say we have the expression seven x squared minus 63. Like always, pause this video, and see if you can factor that. I've intentionally designed
all of these so that you can check whether there's a
common factor across the terms, and here they're all divisible by seven. If you factor out a seven, you're gonna get seven times x-squared minus nine. You might immediately recognize this as a difference of squares. You have x squared minus
this right over here is three squared. Minus three squared. If the term difference of
squares or how to factor them is completely foreign to
you, I encourage you to watch the videos on factoring
difference of squares or do a search on Khan Academy
for difference of squares. But you will see, when you have a difference of squares like this, it can be factored as seven,
this is just a seven out front, and then this part right over
here, get a different color, this part right over here can be written as x plus three times x minus three. It is x squared minus three squared. Now, one thing to appreciate. This really isn't a different technique than the one that we saw
in the previous video. If we just focused on
x squared minus nine, you could view this as x
squared plus zero x minus nine. In that case, you'd say:
"OK, what two numbers "get me a product of negative
nine and add up to zero?" If I need to get a
product of negative nine, that means that they
must be different signs, a positive and a negative; otherwise, if they were the same sign,
you'd get a positive here. So they're different signs. Nine only has three factors. One. You could either have one and nine. There's only two combinations here. You could either have one or
nine, and three and three. And, if you make one
negative or nine negative, that's not going to add up to zero. But if you make one of
these threes negative, that does add up to zero. Say, OK, my two numbers are gonna be negative three and three. So it's gonna be x minus
three times x plus three. Once again, I'm just focusing on what was inside the parentheses
right over here. You'd put that seven out front if we were doing this
exact same expression. But if you recognize it as
a difference of squares, it might happen for you
a little bit faster. Let's do one more example. Let's say that I have two x
squared plus seven x plus three. In general, when my coefficient on the second-degree
term here is not a one, I try to see is there
a common factor here. But, seven isn't divisible
by two, and neither is three. So I can't use the techniques that I used in the last few videos or even over here, where I say: "Oh,
there's a common factor", and get a leading coefficient of one. If you see a situation like that, it's a clue that factoring
by grouping might apply here. Factoring by grouping. On some level, everything
that we've just done now you could view as special
cases of factoring by grouping. Factoring by grouping, you say, OK, can I think of two numbers that
add up to this coefficient. a plus b is equal to seven. a times b, instead of just
saying it needs to equal to three, it actually needs to
be equal to three times this, three times the leading coefficient, the coefficient on the x-squared term. It needs to be equal to three times two. If you think about it, we've
always been doing that. The other examples we gave, the leading coefficient was a one. When you took the constant term,
and multiplied it by a one, you were just saying a times b needs to be equal to that constant term. If you wanna talk about it more generally, it should be a times b
should be the constant term times the leading coefficient. In the introduction to
factoring by grouping we explained why that works. You should never just accept
this as some magic formula. It makes sense for a very
good mathematical reason. But once you accept that, then it's useful to be able
to apply the technique. So can we think of two
numbers that add up to seven and whose product is equal to six? They're going to have to be the same sign since this is a positive value. And, they're gonna be positive
'cause they're the same sign, and if they're adding
up to a positive value, they're both gonna be positive. Well, let's see, one
and six seems to work. One plus six is seven. One times six is six. In factoring by grouping
we rewrite our expression where we break this up
between the a and the b. So I can rewrite this as two x squared plus six x plus, I could write one x. Actually, lemme just do that. Plus one x plus three. As you can see, the
seven x, different color, the seven x has been broken
up into the six x and one x. That whole exercise I
just did is to see how we can break up this first-degree
term right over here. But then, what's useful
about this, is now, we can do the reverse of the
distributive property twice. So, for these first two terms, in a different color than I just use, these first two terms,
you see a common factor. Two x squared and six x,
they're both divisible by two x. So let's factor out a two x
outta those first two terms. If you do that, two x
squared divided by two x, you're just gonna have an x left over. Six x divided by two x, you're
just going to have a three. And then you have plus. Over here, this is a
special situation where, x plus three, there is no common
factor between x and three, so we'll just rewrite that, x plus three. When I put a parentheses on it, which is equivalent to writing
it without a parentheses, you might see something else. I can undistribute, or I can
factor out, an x plus three. What happens if I do that? I'm gonna get an x plus three. And then I'm gonna have
leftover in this term, if I factored out an x plus three, I'm just gonna have a two x left over. Two x. And then, this term, if I
factor out an x plus three, I'm just gonna have a one left over. Plus one. Lemme do it in that same color. Having trouble switching colors today. Two x plus one. And we are done. So, as I said, these are
all various techniques. On some level, factoring by grouping is sometimes viewed as the hardest one. But I'll say hard in parentheses because everything we did is
just a variation, really, a special case, of factoring by grouping. As you can see, it's all about, two numbers that add up
to that middle coefficient on the first-degree term when
it's written in standard form. Their product is equal to the product of the constant and the
leading coefficient. If you do that, you
break it up, it works out quite nicely where you
keep factoring out terms. This one, on some levels,
was a little bit more subtle, 'cause you had to recognize
that this x plus three has a one coefficient on there implicitly. One times x plus three is the
same thing as x plus three. And then see that you can
factor out an x plus three outta both of these terms,
and then, once you do that, you're gonna be left
with the two x plus one. But all of these, if you
really feel comfortable with this arsenal of techniques, you're gonna do pretty well. Frankly, if none of these work, well, you might already be familiar
with the quadratic formula or you might be soon to learn it, but that's when the quadratic
formula might be effective if none of these techniques work.