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### Course: Algebra 1 > Unit 13

Lesson 6: Factoring quadratics by grouping- 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?

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# Factoring quadratics by grouping

Sal factors 4y^2+4y-15 as (2y-3)(2y+5) by grouping. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- i really can not bare the curiosity of why the tecnic Sal uses(in problem 4y^+4y-15,

a+b=4*15)makes sense. why do you have to multiply the product of y^ and 15?

i get that if you multiply that way you can solve the problems but i'm pretty sure that the mathmetician who dicovered this didn't figure out is suddenly with no thought.

i know it works but can someone please tell me WHY it works?(57 votes)- Any equation with a factored form of (ax+b)(cx+d) will multiply, by distribution, to get acx^2 + (ad + bc)x + bd.

You can then multiply the coefficient of x^2 and the constant (ac*bd) like the instructor suggests.

Notice that this is all multiplication a*c*b*d, therefore, using the commutative property, ac*bd=ad*bc.

ad*bc is the product of the two numbers, ad and bc, who's sum is the middle term of the trinomial.(36 votes)

- I still don't get where the minus 60 is coming from...(16 votes)
- For equations in the form
`ax^2 + bx + c`

, you need to find two numbers that multiply together to get`a*c`

and add together to get`b`

. Here`a*c`

=`4*-15`

=`-60`

.

For more see: https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-by-grouping/v/factor-by-grouping-and-factoring-completely(29 votes)

- Can somebody tell me why grouping works? I get that it's factoring, but how do we know to split the middle number into factors of the multiple of the first coefficient and the last constant? Are there any videos or links explaining this?(14 votes)
- When a number is written such that,

(a+x)(b+x)

It can also be factorize as

ab+ax+xb+x^2

as we factorize it we get first factor as ab

and the 2nd and 3rd factor as ax+bx.

So we're kinda just doing the reverse of it for quadratic polynomial like these by finding two number which satisfy both ab and ax+bx.

Hope it helps :D(12 votes)

- for the 4 * 15, is it the A term 4, or the B term 4(11 votes)
- I'm confused about which 4 he used. Was it the one before y^2 or the one before y?(7 votes)
- He used the 4 in front of y^2.(7 votes)

- I don’t really get this. I really need help on the skill “Compound Inequalities”, but I feel like Sal is talking way too fast and I don’t get what he is saying. Help me please!(7 votes)
- May I recommend going back and watching some of the previous videos on factoring?(6 votes)

- so at the part where he had the 2 factors -6 and 10 how did you know which factor to put down first(5 votes)
- The order doesn't matter. The only thing thatt matters is that the 2 terms need to add back to the original value (in this case 4y). I'll reverse the 2 terms and redo the problem in the video so that you can see.

4y^2 + 10y - 6y - 15

2y (2y + 5) - 3 (2y + 5)

(2y + 5)(2y - 3)

These are the same 2 factors that Sal created in the video.

Hope this helps.(9 votes)

- In the video he says to multiply a*b, but uses the numbers for a*c. Simple error when doing the video, but fatal if students try to multiply a*b instead of a*c. I have a student who was watching this video with his father. They pointed out to me that I must be doing something wrong because I keep tell them to multiply a*c. Please check this video for the error.(5 votes)
- He is just using 'a' and 'b' to represent any two numbers. He is not using 'a' and 'b' from 'ax^2 + bx + c' If you look at the transcript or listen to that part of the video again it says, "So we're looking for two numbers whose product-- let's call those a and b-- is going to be equal to 4 times..." He is defining 'a' and 'b' right there as any two numbers that satisfy the conditions he continues to lay out.(4 votes)

- What happens if the 2nd degree constant is a variable?(4 votes)
- Your language is off, the constant is a number without a variable (that is the variable is to the 0th power). I think you mean that the second-degree term does not show a coefficient (his example has a coefficient of 4). If you only see the variable x^2, then there is an invisible coefficient of 1 in front which generally makes factoring easier if possible. So use coefficient as a number in front of a variable and constant as a number without a variable.(5 votes)

## Video transcript

We're asked to factor 4y squared
plus 4y, minus 15. And whenever you have an
expression like this, where you have a non-one coefficient
on the y squared, or on the second degree term-- it could
have been an x squared-- the best way to do this
is by grouping. And to factor by grouping we
need to look for two numbers whose product is equal to
4 times negative 15. So we're looking for two numbers
whose product-- let's call those a and b-- is going
to be equal to 4 times negative 15, or negative 60. And the sum of those two
numbers, a plus b, needs to be equal to this 4 right there. So let's think about all the
factors of negative 60, or 60. And we're looking for ones that
are essentially 4 apart, because the numbers are going
to be of different signs, because their product is
negative, so when you take two numbers of different signs and
you sum them, you kind of view it as the difference of
their absolute values. If that confuses you, don't
worry about it. But this tells you that the
numbers, since they're going to be of different size, their
absolute values are going to be roughly 4 apart. So we could try out things like
5 and 12, 5 and negative 12, because one has
to be negative. If you add these two you get
negative 7, if you did negative 5 and 12 you'd
get positive 7. They're just still
too far apart. What if we tried 6
and negative 10? Then you get a negative 4,
if you added these two. But we want a positive 4, so
let's do negative 6 and 10. Negative 6 plus 10
is positive 4. So those will be our two
numbers, negative 6 and positive 10. Now, what we want to do
is we want to break up this middle term here. The whole point of figuring out
the negative 6 and the 10 is to break up the 4y into
a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as
negative 6y plus 10y, right? Because if you add
those you get 4y. And then the other sides of it,
you have your 4y squared, your 4y squared and then
you have your minus 15. All I did is expand this into
these two numbers as being the coefficients on the y. If you add these, you
get the 4y again. Now, this is where the
grouping comes in. You group the term. Let me do it in a
different color. So if I take these two guys,
what can I factor out of those two guys? Well, there's a common factor,
it looks like there's a common factor of 2y. So if we factor out 2y, we get
2y times 4y squared, divided by 2y is 2y. And then negative 6y divided
by 2y is negative 3. So this group gets factored
into 2y times 2y, minus 3. Now, let's look at this other
group right here. This was the whole point about
breaking it up like this. And in other videos I've
explained why this works. Now here, the greatest
common factor is a 5. So we can factor out a 5, so
this is equal to plus 5 times 10y, divided by 5 is 2y. Negative 15 divided
by five is 3. And so we have 2y times
2y minus 3, plus 5 times 2y minus 3. So now you have two terms,
and 2y minus 3 is a common factor to both. So let's factor out a 2y minus
3, so this is equal to 2y minus 3, times 2y, times
that 2y, plus that 5. There's no magic happening
here, all I did is undistribute the 2y minus 3. I factored it out of
both of these guys. I took it out of the
parentheses. If I distribute it in, you'd get
back to this expression. But we're done, we
factored it. We factored it into two
binomial expressions. 4y squared plus 4y, minus
15 is 2y minus 3, times 2y plus 5.