Lesson: Factoring - Quadratics

Reinforcement Activities

Comment on Factoring - Quadratics

tsedze - December 27, 2016
what is the trick in arranging the factors in the bracket? I always get the factors right but swap them in the bracket. is there any way to prevent that, or the trick will be to know working with "+ -" signs(that is I am guessing). Thanks.
greenlight-admin's picture

greenlight-admin - December 28, 2016

Yes, you need to know how to work with positive and negative numbers.

For example, if you want to factor x² - 2x - 15, then you're looking for two value such that their product is -15 and their sum is -2.

What are those two numbers? Well, they are -5 and 3 (positive 3)

So, we get: x² - 2x - 15 = (x - 5)(x + 3)

Does that help?

Vinodhini - August 29, 2017
Hello Brent,

For the following question, can you please advise how to find S+T from (S+T)^2 when we have inequality sign.

http://gre.myprepclub.com/forum/manhattan-prep-gre-algebra-question-s-2-t-2-1-2st-2339.html

Thanks :)
greenlight-admin's picture

greenlight-admin - August 29, 2017

You bet.
Here's my response: http://gre.myprepclub.com/forum/manhattan-prep-gre-algebra-question-s-2-...

Cheers,
Brent

kmwagner23 - September 18, 2017
For the comparison question A: x^2 + 1 B: 2x - 1 could you set each expression to zero and solve for x to then compare? I did this and got the correct answer but I am unsure if I could apply this technique consistently to achieve a correct answer. Thank you!
greenlight-admin's picture

greenlight-admin - September 18, 2017

Unfortunately, that strategy won't always work for Quantitative Comparison questions. It's just a coincidence that it worked this time.

That said, how did you solve the equation x^2 + 1 = 0?

wsee159 - November 19, 2017
Good Morning,

This question: http://gre.myprepclub.com/forum/x-4-x-5508.html

can we just plug in numbers for x? I looked at your explanation on the link but to me, it just seemed easier to plug in numbers. Is that wrong for this type of question?
greenlight-admin's picture

greenlight-admin - November 19, 2017

Question link: http://gre.myprepclub.com/forum/x-4-x-5508.html

We can certainly plug in numbers, but doing so won't yield a definitive answer.

Sure, we might plug in 4(or more) different values for x and, each time we do so, we'll find that Quantity B is greater than Quantity A. HOWEVER, we will still have doubts about whether we have failed to test a certain x-value that shows us that Quantity B is not greater than Quantity A.

I discuss this drawback starting at 2:50 in the following video: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...

wsee159 - November 19, 2017
Thanks!

Ravi - March 2, 2018
Hi.. can you provide an explanation for this? https://gre.myprepclub.com/forum/the-integer-v-is-greater-than-1-if-v-is-the-square-of-an-i-1849.html

actually i am not sure if i am able to understand the question itself
greenlight-admin's picture

greenlight-admin - March 3, 2018

Hi Ravi,

Here's my step-by-step solution: https://gre.myprepclub.com/forum/the-integer-v-is-greater-than-1-if-v-is...

Cheers,
Brent

Omar Sanchez - July 9, 2018
Quantity A: X(4-X)
Quantity B: 6


When in doubt, plug in numbers. I plugged in 2, 3, 6 and 100 for x in quantity A. Like a doomed marriage, all efforts led to the same dead end, as it is structured for unavoidable negativity.

Quantity B will always be greater than A.

Greetings to all.
greenlight-admin's picture

greenlight-admin - July 9, 2018

Question link: https://gre.myprepclub.com/forum/x-4-x-5508.html

I love your comments, Omar. Very funny!!

Yes, plugging in different values for x will certainly give you a strong idea of which quantity is greater.

Great work!

Cheers,
Brent

suraj16 - October 1, 2018
Quantity A: x(4-x)
Quantity B: 6

Hi Brent,

For the above QC, is the following approach correct?
x(4-x)
4x-x^2 i.e -x^2+4x
Subtracting by -4, therefore
quantity A: -x^2+4x-4
quantity B: 2

-(x^2-4x+4)
-(x-2)^2
-(some +ve or -ve number)^2
after squaring it becomes positive
-(positive number)
negative number

since quantity A will always be negative, quantity B will be greater.

greenlight-admin's picture

greenlight-admin - October 1, 2018

That's a perfect approach!!
Nice work.

Cheers,
Brent

hmark2013 - May 5, 2019
Hi Brent!
For this question https://gre.myprepclub.com/forum/x-1-y-12436.html
x = 1-y

A: x^2 +2xy+y^2
B: x+y

I solved for the value of X+Y by adding Y to X=1-Y and got X+Y=1.
I then knew that x^2 +2xy+y^2 factored out to be (x+Y)^2. I replaced X+Y with 1 to get 1^2, which made answer A and B equal. Is this a valid approach?
greenlight-admin's picture

greenlight-admin - May 6, 2019

That's a perfectly valid approach. Nice work!

Cheers,
Brent

kevin - July 18, 2019
Regarding this question: https://gre.myprepclub.com/forum/x-4-x-5508.html

When you get to the -1(x-2)^2 + 4 vs 6 part, can I just subtract 4 from both sides to get

-1(x-2)^2 vs 2

and then I could say the left side will always be negative and since the right side is always positive, then the answer is B.

greenlight-admin's picture

greenlight-admin - July 19, 2019

Question link: https://gre.myprepclub.com/forum/x-4-x-5508.html

That's perfect reasoning. Nice work!

jango615 - July 14, 2020
I'm a little confused about the factoring of 49x² + 14x + 1.
I thought the product of the square of the first and last value will equal the middle value. Wouldn't it be 7x?
Thanks
greenlight-admin's picture

greenlight-admin - July 15, 2020

Special Product: a² + 2ab + b² = (a + b)(a + b)

Notice that the middle term (2ab) is TWICE the product of a and b.
Also, the first term (a²) is the square of a.
Likewise, the last term (b²) is the square of b.

Given: 49x² + 14x + 1
The first term (49x²) is the square of 7x.
The last term (1) is the square of 1.
The middle term (14x) is TWICE the product of 7x and 1.
So, we can be certain that 49x² + 14x + 1 is a Special Product

As such, we can factor it as follows:
49x² + 14x + 1 = (7x)² + (2)(7x)(1) + (1)²
= (7x + 1)(7x + 1)

Does that help?

ravin654 - November 15, 2020
https://gmatclub.com/forum/if-y-x-5-2-x-1-2-6-then-y-is-least-when-x-202553.html

for this problem why cant we take the equation and simplifiy it to 2x^2-8x+20, simplify that further to x^2-4x+10 then we know that the parabola is pointing upwards. So cant we find the vertice using x= -b/2a, plug in x into the equation and get y as the minimum, knowing that the vertice is the minimum point of the equation
greenlight-admin's picture

greenlight-admin - November 16, 2020

Question link: https://gmatclub.com/forum/if-y-x-5-2-x-1-2-6-then-y-is-least-when-x-202...

That totally works.
So, with x² - 4x + 10, a = 1, b = -4, so -b/2a = -(-4)/(2)(1) = 2
x = 2 is the answer.

Aside: If I encountered but this question on test day, I'd probably just test the 5 answer choices.

mwahabsultan - October 12, 2022
IMPORTANT: If (something)² < 1, then it must be the case that -1 < something < 1
Since we now know that (s + t)² < 1, we can say that -1 < s + t < 1

Hey Brent do we take sq root to make it s + t from (s+t)^2
greenlight-admin's picture

greenlight-admin - October 12, 2022

Something like that.
To be more precise, we can use a property that says √(k²) = |k|

So we can take: (s + t)² < 1
Take the square root of both sides to get: |s + t| < 1
Simplify to get: -1 < s + t < 1

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