Factoring Polynomials With A Binomial GCF

[spaceshowfeature1]

I'm really struggling with problems that have Binomials as their greatest common Factor. For example, I solve the following problem like this: 3(9+7x)-(2-x)(9+7x) and I get -3(2-x)(9+7x), but apparently the answer is (x+1)(9+7x). What the ****? There are also problems like this: z^2(4z-z^3)+7(z^3-4z). I got help on another forum by solving the latter, but I KNOW that I never would have realized that (4z-z^3)=-(z^3-4z) by my own intuition. How can I realize what to do in problems like these? I tried finding practice problems for problems like these to get better, but the only problems are these problems. Can someone please link me some problems like these for Practice? It is really unsatisfying to have little to partial knowledge on solving these equations. Thanks and Good Day.

 

[Deleted member 4993]

I'm really struggling with problems that have Binomials as their greatest common Factor. For example, I solve the following problem like this: 3(9x+7)-(2-x)(9+7x) and I get -3(2-x)(9+7x), but apparently the answer is (x+1)(9+7x). What the ****? There are also problems like this: z^2(4z-z^3)+7(z^3-4z). I got help on another forum by solving the latter, but I KNOW that I never would have realized that (4z-z^3)=-(z^3-4z) by my own intuition. How can I realize what to do in problems like these? I tried finding practice problems for problems like these to get better, but the only problems are these problems. Can someone please link me some problems like these for Practice? It is really unsatisfying to have little to partial knowledge on solving these equations. Thanks and Good Day.

3(9x+7)-(2-x)(9+7x) .... are you sure that this was the given polynomial?

In the first term you have (9x+7) and in the second term you have (9 + 7x)

You do not have a common factor - as posted.

 

[spaceshowfeature1]

3(9x+7)-(2-x)(9+7x) .... are you sure that this was the given polynomial?

In the first term you have (9x+7) and in the second term you have (9 + 7x)

You do not have a common factor - as posted.

Yes I am sure. I was thinking the common factor was (9+7x).

 

[spaceshowfeature1]

3(9x+7)-(2-x)(9+7x) .... are you sure that this was the given polynomial?

In the first term you have (9x+7) and in the second term you have (9 + 7x)

You do not have a common factor - as posted.

Yes that is the problem. Where can i find more problems like these?

 

[spaceshowfeature1]

Reply

3(9x+7)-(2-x)(9+7x) .... are you sure that this was the given polynomial?

In the first term you have (9x+7) and in the second term you have (9 + 7x)

You do not have a common factor - as posted.

The common factor is (9+7x). Where can I find problems like these?

 

[Denis]

3(9x+7)-(2-x)(9+7x) and I get -3(2-x)(9+7x),
but apparently the answer is (x+1)(9+7x).

If the (9x + 7) on left should be (9 + 7x),
then the given answer is CORRECT...

Don't forget that -(2 - x) = -2 + x : OK?

Btw, any reasons you did not reply to your 1st post (Jul.20)?

 

[JeffM]

I'm really struggling with problems that have Binomials as their greatest common Factor. For example, I solve the following problem like this: 3(9x+7)-(2-x)(9+7x) and I get -3(2-x)(9+7x), but apparently the answer is (x+1)(9+7x). What the ****? There are also problems like this: z^2(4z-z^3)+7(z^3-4z). I got help on another forum by solving the latter, but I KNOW that I never would have realized that (4z-z^3)=-(z^3-4z) by my own intuition. How can I realize what to do in problems like these? I tried finding practice problems for problems like these to get better, but the only problems are these problems. Can someone please link me some problems like these for Practice? It is really unsatisfying to have little to partial knowledge on solving these equations. Thanks and Good Day.

Expand if you do not "see" a factoring. And check your work.

\(\displaystyle 3(9x + 7) - (2 - x)(9 + 7x) = 27x + 21 - (18 + 14x - 9x - 7x^2) =\)

\(\displaystyle 27x + 3 - 5x + 7x^2 = 7x^2 + 22x + 3 = (7x + 1)(x + 3).\)

You say that the answer "apparently" is (x + 1)(9 + 7x). Let's check

\(\displaystyle (x + 1)(9 + 7x) = 9x + 7x^2 + 9 + 7x = 7x^2 + 16x + 9 \ne 7x^2 + 22x + 3.\)

So either the answer you were given is wrong or the problem you posted is wrong.

As for your second problem, expand the expression if you do not "see" anything.

\(\displaystyle z^2(4z-z^3)+7(z^3-4z)= 4z^3 - z^5 + 7z^3 - 28z = -\ z^5 + 11z^3 - 28z =\)

\(\displaystyle (-\ z)(z^4 - 11z^2 + 28) = (-\ z)(z^2 - 7)(z^2 - 4) = (-\ z)(z^2 - 7)(z^2 - 4) =\)

\(\displaystyle (-\ z)(z + \sqrt{7})(x - \sqrt{7})(z + 2)(z - 2).\)

Now, honestly, I do not think it terribly deep to perceive that

\(\displaystyle (4z - z^3) = (-\ 1)(z^3 - 4z).\)

But we all overlook things. Simplifying the expansion means that you do not need to see anything.

 

[spaceshowfeature1]

3(9x+7)-(2-x)(9+7x) .... are you sure that this was the given polynomial?

In the first term you have (9x+7) and in the second term you have (9 + 7x)

You do not have a common factor - as posted.

If the (9x + 7) on left should be (9 + 7x),
then the given answer is CORRECT...

Don't forget that -(2 - x) = -2 + x : OK?

Btw, any reasons you did not reply to your 1st post (Jul.20)?

Sorry, I just kinda forgot about it. Thanks for the responce on the last thread!

 

[spaceshowfeature1]

Expand if you do not "see" a factoring. And check your work.

\(\displaystyle 3(9x + 7) - (2 - x)(9 + 7x) = 27x + 21 - (18 + 14x - 9x - 7x^2) =\)

\(\displaystyle 27x + 3 - 5x + 7x^2 = 7x^2 + 22x + 3 = (7x + 1)(x + 3).\)

You say that the answer "apparently" is (x + 1)(9 + 7x). Let's check

\(\displaystyle (x + 1)(9 + 7x) = 9x + 7x^2 + 9 + 7x = 7x^2 + 16x + 9 \ne 7x^2 + 22x + 3.\)

So either the answer you were given is wrong or the problem you posted is wrong.

As for your second problem, expand the expression if you do not "see" anything.

\(\displaystyle z^2(4z-z^3)+7(z^3-4z)= 4z^3 - z^5 + 7z^3 - 28z = -\ z^5 + 11z^3 - 28z =\)

\(\displaystyle (-\ z)(z^4 - 11z^2 + 28) = (-\ z)(z^2 - 7)(z^2 - 4) = (-\ z)(z^2 - 7)(z^2 - 4) =\)

\(\displaystyle (-\ z)(z + \sqrt{7})(x - \sqrt{7})(z + 2)(z - 2).\)

Now, honestly, I do not think it terribly deep to perceive that

\(\displaystyle (4z - z^3) = (-\ 1)(z^3 - 4z).\)

But we all overlook things. Simplifying the expansion means that you do not need to see anything.

Thanks for taking all that time to work out the problem in detail. You must be more passionate then I am... That is also very useful advice. Does anyone know where I can find practice problems similar to these?

 

[mmm4444bot]

… I solve … 3(9x+7)-(2-x)(9+7x) and I get -3(2-x)(9+7x), but apparently the answer is (x+1)(9+7x). What the ****? …

I mentioned (in your first thread) that you need to show your work, if you want to get good help here. (I'm not sure what your thought process was, on the exercise above. I also don't know what instructions came with it, but I'm fairly certain they did not ask you "to solve".) Going forward, we need to see complete exercise statements and your steps.

Please take a moment to read the forum guidelines.

… How can I realize what to do in problems like these? …

By practicing enough to gain the experience you require to recognize how to proceed with given exercises. Don't get hung up over the fact that you're struggling today. Learning math (or any compex system) is a process of making mistakes. When something goes wrong, accept that mistakes are normal. Figure out what's wrong (or get help), then make sure you understand why the mistake occurred, fix it, and move on.

You wrote that you "never would have realized that (4z-z^3)=-(z^3-4z)". That statement is only true today. With enough practice, you will begin to see and think symbolically, so you will recognize forms like (A-B), and you will remember that -(B-A) is equivalent. How will all of this happen? With experience! (It takes time. We learn by doing.)

… I tried finding practice problems … like these … but [I found only the two I posted] …Can someone please link me some problems like these …

It's not always easy to find worksheets online that are tailored to a single variation of some general topic. (Did you ask your instructor for additional resources?)

Try this page at Khan Academy. Actually, you might find that entire section interesting; if you'd like to start at the beginning lesson, use the links on the left margin.

 

[Denis]

...but I KNOW that I never would have realized that (4z-z^3)=-(z^3-4z)

Do you now "see" that right side is really -1*(z^3-4z) ?

 

[spaceshowfeature1]

I mentioned (in your first thread) that you need to show your work, if you want to get good help here. (I'm not sure what your thought process was, on the exercise above. I also don't know what instructions came with it, but I'm fairly certain they did not ask you "to solve".) Going forward, we need to see complete exercise statements and your steps.

Please take a moment to read the forum guidelines.


By practicing enough to gain the experience you require to recognize how to proceed with given exercises. Don't get hung up over the fact that you're struggling today. Learning math (or any compex system) is a process of making mistakes. When something goes wrong, accept that mistakes are normal. Figure out what's wrong (or get help), then make sure you understand why the mistake occurred, fix it, and move on.

You wrote that you "never would have realized that (4z-z^3)=-(z^3-4z)". That statement is only true today. With enough practice, you will begin to see and think symbolically, so you will recognize forms like (A-B), and you will remember that -(B-A) is equivalent. How will all of this happen? With experience! (It takes time. We learn by doing.)


It's not always easy to find worksheets online that are tailored to a single variation of some general topic. (Did you ask your instructor for additional resources?)

Try this page at Khan Academy. Actually, you might find that entire section interesting; if you'd like to start at the beginning lesson, use the links on the left margin.

Thank you for the advice. I'm trying to get used to the forum guidelines, and I will try to follow them better in the future.

 

[spaceshowfeature1]

Do you now "see" that right side is really -1*(z^3-4z) ?

Yes.