Factor By Grouping With Negative Leading Coefficient

[feliz_nyc]

Factor -6x^2 - 48x - 72 by grouping.

Is the first step to factor out -1?

Doing so, I get -(6x^2 + 48x + 72).

However, I see that each term can be divided by the leading coefficient. Perhaps factoring out -6 is better.

Let me see.

-6(x^2 + 8x + 12).

This can be factored without grouping. My answer is:

-6(x + 2)(x + 6)

You say?

 

[Dr.Peterson]

Factor -6x^2 - 48x - 72 by grouping.

Is the first step to factor out -1?

Doing so, I get -(6x^2 + 48x + 72).

However, I see that each term can be divided by the leading coefficient. Perhaps factoring out -6 is better.

Let me see.

-6(x^2 + 8x + 12).

This can be factored without grouping. My answer is:

-6(x + 2)(x + 6)

You say?

Exactly my thinking. I would first see that I want to factor out the negative just to make things easier; but before I did that, I would see that there is a common factor of 6, and instead factor out -6. Then I'd finish by factoring the trinomial, which is easy; and then check by multiplying (distributing or expanding) to make sure I didn't make some silly error like we all do.

 

[Deleted member 4993]

Factor -6x^2 - 48x - 72 by grouping.

Is the first step to factor out -1?

Doing so, I get -(6x^2 + 48x + 72).

However, I see that each term can be divided by the leading coefficient. Perhaps factoring out -6 is better.

Let me see.

-6(x^2 + 8x + 12).

This can be factored without grouping. My answer is:

-6(x + 2)(x + 6)

You say?

Where did this problem come from? Is it a problem for which the answer was worked out in Prof. Leonard's Lecture? In that case - why do you ask "you say ?"..

 

[Otis]

Factor -6x^2 - 48x - 72 by grouping

There are different ways to find the same answers, but, when I'm instructed to use a particular method, then that's what I do.

Otherwise, do you have a question about this exercise?

 

[feliz_nyc]

Where did this problem come from? Is it a problem for which the answer was worked out in Prof. Leonard's Lecture? In that case - why do you ask "you say ?"..

This is not a Leonard problem. This is an unanswered question I found in a Facebook math group.

 

[feliz_nyc]

There are different ways to find the same answers, but, when I'm instructed to use a particular method, then that's what I do.

Otherwise, do you have a question about this exercise?

What if I switch the problem just a little bit?
What factoring -x^2 - 48x - 72? The leading coefficient is -1. What then?

 

[Otis]

What factoring -x^2 - 48x - 72?

You're asking how to factor -x^2 - 48x - 72 by grouping, yes?

Is the first step to factor out -1?

Yes, that's a first step. After checking your work on that step, what is the next step? (Hint: It involves two factors of one coefficient that sum to another coefficient.)

[imath]\;[/imath]

 

[Deleted member 4993]

What if I switch the problem just a little bit?
What factoring -x^2 - 48x - 72? The leading coefficient is -1. What then?

Can you factorize following expressions:

• a^2 + b^2 + 2 * a * b
• x^2 + y^2 - 2 * x * y
• u^2 - v^2

 

[Dr.Peterson]

What if I switch the problem just a little bit?
What factoring -x^2 - 48x - 72? The leading coefficient is -1. What then?

Then you do whatever you have learned to do.

I myself, after trying briefly, used the discriminant to check whether it can be factored. Is that something you've heard of?

A very important thing to know about algebra is that it is easy to make up a problem that is very hard to solve, or at least that can't be solved by the basic methods you first learn. So making up a problem at random can be a dangerous thing to do, and is not helpful for learning. That's one reason a source of good problems, such as a textbook (especially when answers are available) is valuable -- even if you aren't necessarily learning from that book!

 

[AvgStudent]

What if I switch the problem just a little bit?
What factoring -x^2 - 48x - 72? The leading coefficient is -1. What then?

There are many ways to factor as other people have mentioned, but completing the square is my go-to method if the standard guess and check don't work. Do you know how to complete the square?

 

[Steven G]

Factor -6x^2 - 48x - 72 by grouping.

Is the first step to factor out -1?

Doing so, I get -(6x^2 + 48x + 72).

However, I see that each term can be divided by the leading coefficient. Perhaps factoring out -6 is better.

Let me see.

-6(x^2 + 8x + 12).

This can be factored without grouping. My answer is:

-6(x + 2)(x + 6)

You say?

I say that you should multiply out your answer and see if you get the original expression.

 

[JeffM]

Factor -6x^2 - 48x - 72 by grouping.

Is the first step to factor out -1?

Doing so, I get -(6x^2 + 48x + 72).

However, I see that each term can be divided by the leading coefficient. Perhaps factoring out -6 is better.

Let me see.

-6(x^2 + 8x + 12).

This can be factored without grouping. My answer is:

-6(x + 2)(x + 6)

You say?

Yes, it can easily be factored without grouping, but doing so does not help you learn how to use grouping to factor something.

[math]-6x^2 - 48x - 72 = -6(x^2 + 8x + 12) = -6(x^2 + 2x + 6x + 12) =\\ -6\{x(x + 2) + 6(x + 2)\} = - 6(x + 6)(x + 2).[/math]
Personally, my way of factoring a quadratic is to spend at most a minute trying to see if I can find two numbers that add to the coefficient of the x term and multiply to the constant term. If I do not see such a pair quickly, I set up the determinant. I am not sure that I have ever factored a quadratic by grouping. But the purpose of the exercise is to train you with simple examples on how to do factoring by grouping. And I can assure you that I have no interest in memorizing the cubic or quartic formulas.

 

[feliz_nyc]

Yes, it can easily be factored without grouping, but doing so does not help you learn how to use grouping to factor something.

[math]-6x^2 - 48x - 72 = -6(x^2 + 8x + 12) = -6(x^2 + 2x + 6x + 12) =\\ -6\{x(x + 2) + 6(x + 2)\} = - 6(x + 6)(x + 2).[/math]
Personally, my way of factoring a quadratic is to spend at most a minute trying to see if I can find two numbers that add to the coefficient of the x term and multiply to the constant term. If I do not see such a pair quickly, I set up the determinant. I am not sure that I have ever factored a quadratic by grouping. But the purpose of the exercise is to train you with simple examples on how to do factoring by grouping. And I can assure you that I have no interest in memorizing the cubic or quartic formulas.

Hi JeffM,

Thank you for your nice and informative reply. I am currently revisiting intermediate algebra watching Professor Leonard classroom video lessons. In Lecture 6.2 (you can find in on YouTube), Leonard introduces a method called the Diamond Method (not sure if he is the author of this method) for factoring trinomials of the kind ax^2 + bx + c. Please, check out the lesson by Leonard and tell me if you think it's ok for learning how to factor trinomials. So far, it's working out for me. You say?

 

[Steven G]

Feliz,
You have 114 posts and you just signed up. That probably is a record for any forum member. Please understand that all the helpers here just want to help you with your math and not socialize with you or any other poster.

 

[feliz_nyc]

Feliz,
You have 114 posts and you just signed up. That probably is a record for any forum member. Please understand that all the helpers here just want to help you with your math and not socialize with you or any other poster.

Steven,

I totally get it. Moving forward, I will only post threads in terms of math discussion. No more socializing.

 

[topsquark]

Steven,

I totally get it. Moving forward, I will only post threads in terms of math discussion. No more socializing.

Feel free to use the PM system, if you like.

-Dan

 

[JeffM]

Hi JeffM,

Thank you for your nice and informative reply. I am currently revisiting intermediate algebra watching Professor Leonard classroom video lessons. In Lecture 6.2 (you can find in on YouTube), Leonard introduces a method called the Diamond Method (not sure if he is the author of this method) for factoring trinomials of the kind ax^2 + bx + c. Please, check out the lesson by Leonard and tell me if you think it's ok for learning how to factor trinomials. So far, it's working out for me. You say?

I am not sure that I looked at the specific video that you wanted me to look at because you did not give a URL. But I did find a video by someone named Leonard on the “diamond method.” If you look at my prior answer, the video is nothing more than a graphic of what I explained in the first sentence of my second paragraph, which is formally known as the rational root theorem applied to quadratics.

There is a theorem (which is not easy to prove) that says there exists a factoring of any polynomial of degree n into n linear binomials. (The proof of this Fundamental Theorem of Algebra is surprisingly hard.) The rational root theorem is guaranteed to provide that factoring if a rational factoring exists, but it does not guarantee that such a factoring does exist. So, we know that (1) it is always possible to factor a quadratic into two linear terms, (2) it is always possible to do so using the diamond method if a rational factorization exists, and (3) there is no guarantee that a rational factorization does exist. In short, the diamond method is looking for a needle in a haystack before you know that there is any needle. Way too much time is wasted on this topic. Unless you “see” it almost immediately, use the discriminant method as far speedier and less error prone.

Moreover, if the “diamond method“ works for a quadratic (and remember it may not), you do not need the grouping method at all. The grouping method is more valuable for polynomials of higher degree than quadratics.

To sum up

1: the diamond method does not work for all quadratics and is no help for polynomials of degree higher than 2;

2: applying the grouping method to quadratics has no practical value othen getting familiar with the method;

3: the discrimanent method always works for quadratics;

4: for polynomials higher in degree than a quadratic, the rational root method is not guaranteed to work, but will always work whenever a rational factoring is possible; and

5: the grouping method will always work in principle, but in practice it may be so inefficient that the universe will end before you get the answer.

Factoring quadratics is easy. Factoring polynomials of degree above 3 may be very hard.

 

[feliz_nyc]

I am not sure that I looked at the specific video that you wanted me to look at because you did not give a URL. But I did find a video by someone named Leonard on the “diamond method.” If you look at my prior answer, the video is nothing more than a graphic of what I explained in the first sentence of my second paragraph, which is formally known as the rational root theorem applied to quadratics.

There is a theorem (which is not easy to prove) that says there exists a factoring of any polynomial of degree n into n linear binomials. (The proof of this Fundamental Theorem of Algebra is surprisingly hard.) The rational root theorem is guaranteed to provide that factoring if a rational factoring exists, but it does not guarantee that such a factoring does exist. So, we know that (1) it is always possible to factor a quadratic into two linear terms, (2) it is always possible to do so using the diamond method if a rational factorization exists, and (3) there is no guarantee that a rational factorization does exist. In short, the diamond method is looking for a needle in a haystack before you know that there is any needle. Way too much time is wasted on this topic. Unless you “see” it almost immediately, use the discriminant method as far speedier and less error prone.

Moreover, if the “diamond method“ works for a quadratic (and remember it may not), you do not need the grouping method at all. The grouping method is more valuable for polynomials of higher degree than quadratics.

To sum up

1: the diamond method does not work for all quadratics and is no help for polynomials of degree higher than 2;

2: applying the grouping method to quadratics has no practical value othen getting familiar with the method;

3: the discrimanent method always works for quadratics;

4: for polynomials higher in degree than a quadratic, the rational root method is not guaranteed to work, but will always work whenever a rational factoring is possible; and

5: the grouping method will always work in principle, but in practice it may be so inefficient that the universe will end before you get the answer.

Factoring quadratics is easy. Factoring polynomials of degree above 3 may be very hard.

A great detailed reply.