Factoring

by on October 1st, 2009
Jake is currently the Academic Director at Grockit, and is located in San Francisco, CA. Visit Grockit for more test prep advice.

Factoring can be seen as reverse-multiplication. In essence, it’s division without actually dividing. Knowing when and how to efficiently rearrange mathematical expressions and/or to reduce larger numbers into more manageable ones will prove invaluable when taking a timed test.

No Long Division

Only in rare cases (such as finding a repeating remainder without a calculator) might we use long division. The vast majority of the time, we will try to reduce numerator and/or denominator of the fraction and look for common factors. Note: 99% of the time you will want to divide before multiplying out. For example:

For what values is 14x(3x + 9) / (2x + 6) < 42 ? x <> -3

Instead of multiplying out (or cross multiplying), we first simplify the fraction in order to eliminate common factors. (Note that there are two powers of x on the top and only one power of x on the bottom.) Even numbers are always a good place to start. If you see both top and bottom contain a factor of two, then you can cancel that first. Then, factor out common multiples (including variables.)

{14x*3(x + 3)}/{2(x + 3)} < 42

7x*3 < 42

21x<42

x<2

When x < 2, the equation {14x(3x + 9)}/{2x + 6} < 42 is satisfied. Pick an easy number to check, like 0.

Common Expressions

Learn these by heart. They come up on every test, in one form or another:

  • x^2 - y^2 = (x - y)(x + y)
  • x^2 + y^2 + 2xy = (x + y)^2
  • x^2 + y^2 - 2xy = (x - y)^2

Note the difference in the last two. Only the sign on the 2xy term changes. So, in the following example, we can recognize what the expressions would be if we were to multiply out.

What is the value of xy?

(x + y)^2 = 9

(x - y)^2 = 16

Note that the question is asking for xy, and not necessarily the value of x or y independently. This should be hint that we don’t need to solve for x and y, and then go about finding xy. Instead we subtract the equations.

x^2 + y^2 + 2xy = 9

x^2 + y^2 - 2xy = 16

(x^2 + y^2 + 2xy) - (x^2 + y^2- 2xy) = 9 - 16

4xy = -7

xy = -7/4

Pulling Perfect Squares Out of Square Roots

When confronted with a large number or expression under a square root sign, remember that you can factor perfect squares out of the expression and “pull” them out from under the square root. This can work with numbers AND variables.

sqrt{(12)(21)x^6 - (10)(18)x^6} =

A. 6sqrt{7x}

B. 6x^3sqrt{2}

C. 24x^6sqrt{3}

D. 6sqrt{7x^3} - 6sqrt{5x^3}

E. 36x^6

Instead of multiplying out and then combining the two terms, and THEN, finding the square root of that equation, we can find common perfect square and take those out first.

Both 12*21 and 10*18 have two factors of two (4) and two factors of three (9).

Since x^6 is also a perfect square, we can rewrite the expression:

sqrt{36x^6(7-5)} = 6x^3sqrt{7-5} = 6x^3sqrt{2}

Choice B.

Take Hints from Other Expressions in the Question

On questions involving factoring, there are occasions on which you will run into expressions that can guide your thinking. In the following example, both the expression in the question and the expression in Statement 1 can be factored similarly.

If y > 0″ title=”y > 0″/>, is <img src= divisible by 4?

(1) y^2 + y is divisible by 10

(2) For a certain integer k, y = 2k + 1

If we break y^3 - y into y(y + 1)(y - 1), we can quickly see that the three factors are consecutive numbers. In order for the product of three consecutive numbers to be divisible by 4, either one has to be divisible by 4 by itself, or two must be divisible by 2 individually. This means that both (y - 1) and (y + 1) must be even.

Statement (1) tells us that y(y + 1) is divisible by 10, which means that one of the two factors is divisible by 5 and also that one of the two is divisible by 2. However, it does not tell us which of the two is divisible by 2, so we cannot know if the remaining (y - 1) is also divisible by 2.

Statement (2) tells us that y = 2k + 1. Since 2k is even, that means 2k + 1 is odd. If y = 2k + 1, then y is also odd. (y - 1) and (y + 1) must both be even, so their product is divisible by 4.

Choice B.

Any other helpful factoring tips? Feel free to leave questions or comments in the comments field below!

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3 comments

  • Nee on October 2nd, 2009 at 10:56 am

    In the last example, I did not understand how you came to the answer B. It should be D according to me- Here is my solution-

    y^3 - y = y(y + 1)(y - 1)
    From 1-> y^2 + y = 10n (n being any random integer)
    From 2-> y = 2k + 1 => y-1 =2k

    Now, putting the value of y-1 from 2 into the main equation we get:-
    y^3 - y = y(y+1)2k = 2k(y^2 + y)
    Now putting the value of (y^2 + y)from 1 into the equation above we get:-
    y^3 - y = 20kn
    This is clearly divisible by 4 and hence the answer is D

    Can you please explain your answer!!

    Reply to this comment

  • Jake Becker on October 2nd, 2009 at 11:26 am

    Nee,

    In your process, you used both Statements 1 and 2. Namely, you got y(y+1) = 10n from (1) and (y-1) from (2). Answer choice (D) refers to each alone being sufficient, so the fact that you use both (1) and (2) eliminate that option.

    (1) is insufficient, because it can be satisfied in 2 ways: One way such that y(y+1)(y-1) is divisible by 4, and the other where it is not. For example:

    y=9, y+1=10, y-1=8 --- In this scenario, (1) is satisfied and the product is divisible by 8.

    y=10, y+1=11, y-1=9 --- In this scenario, y(y+1) is divisible by 10, but y(y+1)(y-1) is not divisble by 4.

    (2) ALONE is sufficient (choice B) because it necessitates that 2 of the 3 factors are even, therefore the product is divisible by 4.

    Reply to this comment

  • Nee on October 4th, 2009 at 10:56 am

    Hi Jake, thanks for the explanation. I got it now. B is indeed the correct answer.

    Reply to this comment

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