Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 25 Mar 2017, 06:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Third Degree Equations

Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7246
Location: Pune, India
Followers: 2204

Kudos [?]: 14326 [2] , given: 222

### Show Tags

10 Sep 2013, 20:50
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
Here is a post on solving third degree equations:

How do you solve an equation such as $$x^3 - 6x^2 + 11x - 6 = 0$$?
How do we get the values of x which satisfy this equation?

If you do get a third degree equation, it will have one very easy root such as 0 or 1 or -1 or 2 or -2 etc. Try a few of these values to get the first root. Here x = 1 works. It is easy to see since you have two 6s, a 1 and an 11 as the coefficients.
Putting x = 1: $$(1)^3 - 6(1)^2 + 11(1) - 6 = 0$$
So you know that (x – 1) is a factor. Now figure out the quadratic which when multiplied by (x – 1) gives the third degree expression

$$(x - 1)(ax^2 + bx + c) = x^3 - 6x^2 + 11x - 6$$
How do we figure out the values of a, b and c? Let’s see.
Coefficient of x^3 on right hand side is 1. So you know that all you need is x^2 so that it multiplies with x to give x^3 on left hand side too. So ‘a’ must be 1.

$$(x - 1)(x^2 + bx + c) = x^3 - 6x^2 + 11x - 6$$
The constant term, ‘c’, is easy to figure out too. It should multiply with -1 to give -6 on right hand side. Hence c must be 6.

$$(x - 1)(x^2 + bx + 6) = x^3 - 6x^2 + 11x - 6$$
Getting the middle term is slightly more complicated. bx multiplies with x to give x^2 term and you also get the x^2 term by multiplying -1 with x^2. You have -6x^2 on right hand side so you need the same on the left hand side too. You already have -x^2 (by multiplying -1 with x^2) so you need another -5x^2 from bx^2. So b must be -5.

$$(x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6$$
Now you just factorize the quadratic in the usual way. Let’s see how exactly we would do it using a question.

Question: Is $$x^3 + 2x^2 - 5x - 6 < 0$$
Statement 1: -3 < x <= -1
Statement 2: -1 <= x < 2

Solution: We know how to deal with inequalities with multiple factors (discussed http://www.veritasprep.com/blog/2012/06 ... e-factors/).
But the inequality is not split into factors here so we will have to do it on our own.
Let’s first try to find the simple root that this expression must have. Try x = 1, -1 etc. We see that when we put x = -1, the expression becomes 0.
$$x^3 + 2x^2 - 5x - 6 = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = 0$$
So the first factor we get is (x + 1).

$$(x + 1)(ax^2 + bx + c) = x^3 + 2x^2 - 5x - 6$$
a must be 1 since we have $$x^3$$ on the right hand side.
c must be -6 since we have -6 on the right hand side.

$$(x + 1)(x^2 + bx - 6) = x^3 + 2x^2 - 5x - 6$$
$$bx^2 + x^2 = 2x^2$$
So b must be 1.

We get: $$(x + 1)(x^2 + x - 6)$$ which is equal to $$(x + 1)(x + 3)(x - 2)$$ (after splitting the quadratic)
So the question becomes:
Is $$(x + 1)(x + 3)(x - 2) < 0?$$
We already know how to deal with inequalities with multiple factors. The transition points here will be -3, -1 and 2. The expression will be negative in the ranges -1 < x < 2 and x < -3

Statement 1: -3 < x <= -1
In this region, the expression is positive (when -3 < x < -1) or 0 (when x = -1). Hence it will certainly not be negative. This is sufficient to answer the question with ‘No’. Hence statement 1 is sufficient to answer the question.

Statement 2: -1 <= x < 2
In this region, the expression is negative (when -1 < x < 2) or 0 (when x = -1). We cannot say for certain whether it will be negative or not. Hence statement 2 alone is not sufficient to answer the question.

It’s not hard to deal with third degree equations. All you have to do is bring it down to second degree by figuring out one root and then the problem is in a format you already know.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 09 Jul 2013 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 7 Re: Third Degree Equations [#permalink] ### Show Tags 13 Sep 2013, 00:35 Is it possible to use plug-in approach to quickly solve similar questions? Statement 1: -3 < x <= -1 Plug-in: -3,-2,-1 Statement 2: -1 <= x < 2 Plug-in: -1,0,1,2 I understand the range is important in such questions, but then ain't there methods to plug-in and reverse solve? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7246 Location: Pune, India Followers: 2204 Kudos [?]: 14326 [0], given: 222 Re: Third Degree Equations [#permalink] ### Show Tags 13 Sep 2013, 03:58 Disillusioned wrote: Is it possible to use plug-in approach to quickly solve similar questions? Statement 1: -3 < x <= -1 Plug-in: -3,-2,-1 Statement 2: -1 <= x < 2 Plug-in: -1,0,1,2 I understand the range is important in such questions, but then ain't there methods to plug-in and reverse solve? Sure you can but there will be risk since you certainly cannot plug in all values to check. e.g. Imagine that you have a third degree expression which factors out as: Is (x + 1)(x + 2)(x - 3) <= 0? Say, given range is -2 < x < 3 You try -1, 0, 1, 2 All of them satisfy the inequality. Does it mean that in this range, (x + 1)(x + 2)(x - 3) is negative or 0? Does it mean that this statement alone is sufficient? No. because in -2 < x< -1, (x + 1)(x + 2)(x - 3) is positive. While in -1 < x < 3, the expression is negative. So I like to use plug in only when I don't see any other easy way out. As for questions which use plug in to reverse solve, the higher level questions nowadays do not allow that luxury. They are made keeping in mind that you know how to plug in. Only the lower level questions might be solvable as such. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 14405
Followers: 604

Kudos [?]: 174 [0], given: 0

### Show Tags

25 Oct 2016, 03:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Third Degree Equations   [#permalink] 25 Oct 2016, 03:45
Similar topics Replies Last post
Similar
Topics:
3 equation 3 17 Apr 2013, 21:16
Equations. 2 19 Dec 2011, 19:43
equations 8 27 Mar 2011, 08:14
Second degree equation 2 22 Jan 2011, 17:05
3 Equating coefficients in a quadratic equation 5 29 Oct 2010, 06:30
Display posts from previous: Sort by