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### Course: Algebra 1 (Eureka Math/EngageNY)>Unit 4

Lesson 5: Topic A: Lessons 3-4: Factoring by grouping

# Strategy in factoring quadratics (part 2 of 2)

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?

• At , how did he get the +1 to add to the equation when factoring out (x+3)?
• 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.
• Lets just pull a zero out of no where... ugh.
• 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.
• 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
• 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.
• 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
• 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!
• What about equations like V^4+v^4x^4?
• Do you mean that you would have an exponent exponented? I don't quite follow you.
• 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,
• 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).
• How is factoring using for life?
• Is there a way to solve the a + b = # and a * b = # without guessing and checking?
(1 vote)
• You would have a system of equations to solve. When you tried to solve it, you would get a quadratic equation to solve.

To make this easier / faster, make sure you know the multiplications tables up thru at least 12. And, learn the divisibility tests so that you can find possible factors to consider. You can find the lessons for divisibility tests by using the search bar at the top of any KhanAcademy page.
• Something mysterious is going on between and :
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.