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## College Algebra

### Course: College Algebra>Unit 4

Lesson 4: 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?
• 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.
• I still don't get where the minus 60 is coming from...
• 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?
• 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
• for the 4 * 15, is it the A term 4, or the B term 4
• I'm confused about which 4 he used. Was it the one before y^2 or the one before y?
• He used the 4 in front of y^2.
• 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!
• May I recommend going back and watching some of the previous videos on factoring?
• so at the part where he had the 2 factors -6 and 10 how did you know which factor to put down first
• 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.
• this is hard
(1 vote)
• no skill issue
• What happens if the 2nd degree constant is a variable?
• 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.
• I'm confused on . Since it should be a+b, that would mean 4+(-15)= -11 and not 4. I know he got 4 from 4y, but that is not the value of the sum from a+b. In the future, do I disregard the sum of a+b and just find the value of the coefficient as the answer for a+b?
(1 vote)
• Sal uses "a" and "b" to represent the two factors he needs to find. They do not refer to any of the coefficients or constant terms in the original quadratic.

To use grouping to factor, you multiply the coefficient of X^2 (4) with the constant term (-15).
4(-15) = -60

Next, you find two numbers that multiply to -60 and also add to the middle term 4. This is where Sal gets:
a*b = -60
a+b = 4

The two numbers that work are -6 and 10.

Hope this helps.