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### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 1

Lesson 2: Topic B: Lessons 12-13: Factorization- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Factor polynomials: quadratic methods
- Factoring two-variable quadratics
- Factoring two-variable quadratics: rearranging
- Factoring two-variable quadratics: grouping
- Factor polynomials: quadratic methods (challenge)
- Factoring using the perfect square pattern
- Factoring using the difference of squares pattern
- Factoring difference of squares: two variables (example 2)
- Factor polynomials using structure
- Factoring higher-degree polynomials
- Factoring sum of cubes
- Factoring difference of cubes
- Finding zeros of polynomials (1 of 2)
- Finding zeros of polynomials (2 of 2)
- Finding zeros of polynomials (example 2)
- Zeros of polynomials (with factoring)

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# Factoring quadratics: common factor + grouping

Sal factors 35k^2+100k-15 as 5(k+3)(7k-1). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Are polynomials the same as quadracts?(11 votes)
- Quadratics are a special kind of polynomial. Here are some examples of various kinds of polynomials:

(1) x^2 + 3x + 9

(2) x^3 + x^2 - 9x

(3) x^5 - 5x^3 - 2x^2 + x - 20

(4) x^10 + x - 1

While each of the above is a polynomial, only (1) is called a quadratic -- this is because its largest exponent is a 2. Another way of saying this is that (1) is a "second-degree polynomial". (2) is a "third-degree polynomial" because its largest exponent is 3 (these kinds of polynomials are often called "cubics"). (3) is a "fifth-degree polynomial", (4) would be a "tenth-degree polynomial", and so on.

So basically, the term "quadratic" is just another word for any second-degree polynomial -- or any polynomial where the largest exponent is 2.

(Also, I used x as my variable in all the above examples, but you could use any variable you like... it doesn't need to be x)(61 votes)

- Okay, so we found out that this factors into 5(7k-1)(k+3), but what does k equal?

How do you find out what k is when there's a 5 outside the factored terms?(11 votes)- Scotty,

The original problem was

Factor 35k² + 100k - 15

This is an expression, but it is not an equation. So you cannot solve the equation for k. All you can do is factor the expression.

If instead the original problem was

35k² + 100k - 15 = 0 you could slove for k

You would first factor the left side, just like in the video and you would get

5(7k-1)(k+3) = 0

The expression on the right would be equal to zero when either (7k-1) = 0 or if (k+3) = 0 because anythng times zero is zero.

And the solutions for k would be k=-3 and k= 1/7

But that was not the solution to the expression in the video because it was not an equaltiy, all we could do is factor the expression.

Factoring an expression may not seem like it has a purpose until you can then use it to solve something, but learning how to factor expressions is the first step to solving quadratic equations.

I hope that helps you make some sense of things.(46 votes)

- Hi at1:03-1:06, Sal mentions finding a number whose product is 7 * -3. Why isn't he simply referring to the number -3 as his product?

Thanks in advance.(11 votes)- The product of two numbers is the number that results from multiplying two numbers together. For example, -3×7 equals -21. In this instance, -21 is the product.

I hope this helps!(4 votes)

- is this the same as factorising using common prime factors?(7 votes)
- Mmm not really - I don't see what the connection is between prime factoring a number and this.(6 votes)

- I understand the concept, but I have a problem that doesn't make sense. If the highest exponent variable
**doesn't**have a constant before it, how would I find the common factor?

The problem is:`y^4 + 10y^2 + 24=___`

Thanks in advance(3 votes)- For now, ignore the first term and focus on the last two. Ask yourself what adds to 10 and multiplies to 24? The correct answer is 6 and 4. So the factored form of this expression would be (y^2+6x)(y^2+4x)(10 votes)

- Would using brackets be acceptable?(4 votes)
- Are there by any chance any videos explaining how to factor problems like:

2x^3 - 16x^2 + 24x

If so, I would appreciate if someone could provide me a link to where I can learn how to solve these types of problems, or teach me themselves.

Thanks a bunches!(1 vote)- 2x³ - 16x² + 24x

Step 1: Factor out the GCF, in this case it is 2x

2x[x² - 8x + 12]

Step 2; Factor [x² - 8x + 12]

(x-6)(x-2)

Step 3: put it all together

(2x)(x-6)(x-2)

If you need to know how to factor cubic equations that don't have a GCF to factor out that makes it easy, let me know and I can walk you through that.(12 votes)

- Consider the polynomial function

Px=x4-3x3+ax2-6x+14,

where a is an unknown real number. If x-2is a factor of this polynomial, what is the value of a?(3 votes)- Interesting question.

Well since x - 2 is a factor of P(x), we know that 2 is a root of P(x). So we must have P(2) = 0. This means:

2⁴ - 3(2³) + 4a - 12 + 14 = 0

Now we can just solve for "a":

16 - 24 + 4a + 2 = 0

8 - 12 + 2a + 1 = 0

-3 + 2a = 0

2a = 3

a = 2/3

Comment if you have any questions.(7 votes)

- why do the questions have upvote and downvote buttons like Reddit?(1 vote)
- So people can upvote the good questions so that other people with the same question can just read the response. The answers also have upvotes and downvotes so people can upvote the best answer.(9 votes)

- Does it matter, in this example, if the 21k went first or the -k would. When I practice, I have no idea which one I should put first and sometimes get it wrong because of that. I'm really confused, please help!(3 votes)
- It can matter. It's not wrong to put the -1k first, but sometimes you'll get stuck trying to factor like that.

You want to think about the idea of "grouping" - which term groups better with it's neighbor. So 21k would be better to group with 7 k^2 because 21 is divisible by 7.

In this example, 21 is also divisible by 3, so you can group the other way:

5(7k^2 -1k + 21k - 3)

5[k(7k-1) + 3(7k-1)]

5(7k-1)(k+3)

Here you get the same answer, but it might not always factor as neatly both ways.

[Hmm, looking at the examples from the videos again, it seems like it**would**factor either way. Does anyone have an example where it doesn't? Or know the rule for it?](4 votes)

## Video transcript

We're asked to factor 35k
squared plus 100k, minus 15. And because we have a non-1
coefficient out here, the best thing to do is probably to
factor this by grouping. But before we even do that,
let's see if there's a common factor across all of these
terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient,
we'll at least have a lower coefficient here. And if we look at all of these
numbers, they all look divisible by 5. In fact their greatest
common factor is 5. So let's at least
factor out a 5. So this is equal to 5
times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided
by 5 is negative 3. So we were able to factor out a
5, but we still don't have a 1 coefficient here, so we're
still going to have to factor by grouping. But at least the numbers here
are smaller so it'll be easier to think about it in terms of
finding numbers whose product is equal to 7 times negative
3, and whose sum is equal to 20. So let's think about that. Let's figure out two numbers
that if I were to add them, or even better if I were to take
their product, I get 7 times negative 3, which is equal
to negative 21. And if I were to take their
sum, if I add those two numbers, it needs to
be equal to 20. Now, once again, because their
product is a negative number, that means they have to be of
different signs, so when you add numbers of different signs,
you could view it as you're taking the difference
of the positive versions. So the difference between the
positive versions of the number has to be 20. So the number that immediately
jumps out is we're probably going to be dealing with 20
and 21, and 1 will be the negative, because we want
to get to a positive 20. So let's think about it. So if we think of 20 and
negative 1, their product is negative 21. Sorry. If we take 21 and negative 1,
their product is negative 21. 21 times negative 1
is negative 21. and if you take their sum,
21 plus negative 1, that is equal to 20. So these two numbers right
there fit the bill. Now, let's break up this 20k
right here into a 21k and a negative 1k. So let's do that. So let's rewrite the
whole thing. We have 5 times 7k squared, and
I'm going to break this 20k into a-- let me do it in
this color right here-- I'm going to break that 20k into
a plus 21k, minus k. Or you could say minus
1k if you want. I'm using those two factors
to break it up. And then we finally have the
minus 3 right there. Now, the whole point of doing
that is so that we can now factor each of the two groups. This could be our first
group right here. And so what can we factor out
of that group right there? Well, both of these are
divisible by 7k, so we can write this as 7k times-- 7k
squared divided by 7k, you're just going to have
a k left over. And then plus 21k divided by
7k is just going to be a 3. So that factors into that. And then we can look at
this group right here. They have a common factor. Well, we can factor out a
negative 1 if we like, so this is equal to negative 1 times--
k divided by negative 1 is k. Negative 3 divided by negative
1 is positive 3. And, of course, we have this
5 sitting out there. Now, ignoring that 5 for a
second, you see that both of these inside terms have
k plus 3 as a factor. So we can factor that out. So let's ignore this
5 for a second. This inside part right here,
the stuff that's inside the parentheses, we can factor k
plus 3 out, and it becomes k plus 3, times k plus 3,
times 7k minus 1. And if this seems a little
bizarre to you, distribute the k plus 3 on to this. K plus 3 times 7k is that term,
k plus 3 times negative 1 is that term. And, of course, the whole
time you have that 5 sitting outside. You have that 5. We don't even have to put
parentheses there. 5 times k plus 3, times
7k minus 1. And we factored it,
we're done.