Factoring Polynomials with a Leading Coefficient Greater than 1 with negatives help?

[Iceycold12]

Hello all. I'm new here and was looking for a math help forum, there was one similiar to this forum one but it seems to have gotten closed.

So my class is doing Factoring Polynimials with a Leading Coefficient greater than 1. I know it very well, but I have an issue with a particular scenario in which negatives are involved.

So here's an example: (Will explain the best I can)

10x^2+31x-14
1. Multiplying 10 x -14=-140
2. Need to find two numbers that when multiplied add to -140 and when added add to 31
3. Numbers are 35 & -4. 35+-4=31 & 35 x -4 = -140.
4. Rewriting equation: 10x^2+35x-4x-14
5. Grouping. (10x^2)-(4x-14) <- This part confuses me, my teacher told me to re-write it as -(4x+14) because if the negative was distributed it would make them both negative. And that is in the original equation 10x^2+35x-4x-14.

6. SO when factoring I know it needs to be a negative. So what do I do, factor that part of the equation as: (10x^2+35x)-(4x+14) as -> 5x(2x+7) + -2(2x+7)?

Given the above is right then can it be said that the part -(4x+14) can be treated as -(-4x-14) (the original one) and factor that? How would it be done?

OR can I factor it like this: -(4x+14) which would be 2(2x+7) and make the 2 outside the parenthesis negative?

ALSO. What would happen if the problem was switched in grouping for example. Instead of what I have there it'd be: 2k^2+3k-16k-24

Tried to explain it my best. Hope you understood.

Essentially I get confused when this is the scenario: (numbers) negative sign (numbers)

Sorry for the lengthy post but with Algebra if I don't understand something I feel I will suffer in the next lesson.

Thanks!

 

[pka]

FACTOR
10x^2+31x-14

First step find prime factors: \(\displaystyle 10=2\cdot 5~\&~14=2\cdot 7\).
Now look for how to get \(\displaystyle 31\).
Well \(\displaystyle 5\cdot 7-2\cdot 2\) does it.

 

[Deleted member 4993]

Hello all. I'm new here and was looking for a math help forum, there was one similiar to this forum one but it seems to have gotten closed.

So my class is doing Factoring Polynimials with a Leading Coefficient greater than 1. I know it very well, but I have an issue with a particular scenario in which negatives are involved.

So here's an example: (Will explain the best I can)

10x^2+31x-14
1. Multiplying 10 x -14=-140
2. Need to find two numbers that when multiplied add to -140 and when added add to 31
3. Numbers are 35 & -4. 35+-4=31 & 35 x -4 = -140.
4. Rewriting equation: 10x^2+35x-4x-14
5. Grouping. (10x^2)-(4x-14) <- This part confuses me, my teacher told me to re-write it as -(4x+14) because if the negative was distributed it would make them both negative. And that is in the original equation 10x^2+35x-4x-14.

6. SO when factoring I know it needs to be a negative. So what do I do, factor that part of the equation as: (10x^2+35x)-(4x+14) as -> 5x(2x+7) + -2(2x+7)?

You should write the above as:

5x(2x+7) - 2(2x+7)

= (2x + 7) * (5x -2)



Given the above is right then can it be said that the part -(4x+14) can be treated as -(-4x-14) (the original one) and factor that? How would it be done?

OR can I factor it like this: -(4x+14) which would be 2(2x+7) and make the 2 outside the parenthesis negative?

ALSO. What would happen if the problem was swithced in grouping for example. instead of what I have there it'd be: 2k^2+3k-16k-24

It really does not matter how you group those - the final answer would same.

say you did:

10x² - 4x + 35x -14

= 2x (5x - 2) + 7(5x - 2)

= (5x-2)*(2x+7) ..................see the answer is same.



Tried to explain it my best. Hope you understood.

Essentially I get confused when this is the scenario: (numbers) negative sign (numbers)

Sorry for the lengthy post but with Algebra if I don't understand something I feel I will suffer in the next lesson.

Thanks!

Thanks for showing detailed work - it saves us lot of time and effort.

 

[Iceycold12]

Yeah I figured for you guys to see what I mean I'd have it to explain it in detail.

Maybe I'm overlooking something here but I still quite don't understand. I'm gonna grab a portion of the problem.

10x^2+35x-4x-14

Tell me if I'm understanding this correctly.

When this is put into parenthesis the negative is "cut out". So: (10x^2+35x)-(4x-14) My teacher though told me to put -(4x+14), because if you were to distribute this, then it would equal to the original equation of -4x-14. That's where I'm totally puzzled. So do I factor out a negative number or a positive?

 

[Iceycold12]

Yeah I figured for you guys to see what I mean I'd have it to explain it in detail.

Maybe I'm overlooking something here but I still quite don't understand. I'm gonna grab a portion of the problem.

10x^2+35x-4x-14

Tell me if I'm understanding this correctly.

When this is put into parenthesis the negative is "cut out". So: (10x^2+35x)-(4x-14) My teacher though told me to put -(4x+14), because if you were to distribute this, then it would equal to the original equation of -4x-14. That's where I'm totally puzzled. So do I factor out a negative number or a positive?

What would happen in this scenario: 2k^2+3k-16k-24 Would the 16k and 24 become positive and be factored with a positive number?

 

[pappus]

Hello all. I'm new here and was looking for a math help forum, there was one similiar to this forum one but it seems to have gotten closed.....

You find this forum now under MathHelpBoards.com. Have a look!

...
What would happen in this scenario: 2k^2+3k-16k-24 Would the 16k and 24 become positive and be factored with a positive number?

If you group the given term like this:

2k^2+3k-16k-24 = 2k^2-16k+3k-24 = 2k(k-8)+3(k-8)

and then factor out (k - 8)

 

[Iceycold12]

You find this forum now under MathHelpBoards.com. Have a look!



If you group the given term like this:

2k^2+3k-16k-24 = 2k^2-16k+3k-24 = 2k(k-8)+3(k-8)

and then factor out (k - 8)

So you can just re-group them so that you don't face a (number)-(number) situation? When you do encounter those, I'm still relatively confused on what to do. Factor a negative, or a positive? -(4x+14)

Thanks.

 

[pka]

I'm overlooking something[/COLOR] here but I still quite don't understand. I'm gonna grab a portion of the problem. 10x^2+35x-4x-14

Big time
\(\displaystyle 10x^2+35x-4x-14= 10x^2+31x-14=(5x-2)(2x+7)\)

 

[lookagain]

First step find prime factors: \(\displaystyle 10=2\cdot 5~\&~14=2\cdot 7\).
Now look for how to get \(\displaystyle 31\).
Well \(\displaystyle 5\cdot 7-2\cdot 2\) does it.

That is not sufficient. You have to consider all of the divisors,
not just the prime factors.

For instance in \(\displaystyle 10x^2 + 139x - 14,\) if you were to only consider the
prime factors of 10 and 14, then you wouldn't come up with the correct
factorization of (10x - 1)(x + 14).

 

[lookagain]

10x^2+35x-4x-14

Tell me if I'm understanding this correctly.

When this is put into parenthesis the negative is "cut out". So: (10x^2+35x)-(4x-14)
My teacher though told me to put -(4x+14), because if you were to distribute this,
then it would equal to the original equation of -4x-14. That's where I'm totally
puzzled. So do I factor out a negative number or a positive?

What would happen in this scenario: 2k^2+3k-16k-24 Would the 16k and 24
become positive and be factored with a positive number?

So you can just re-group them so that you don't face a (number)-(number) situation?
When you do encounter those, I'm still relatively confused on what to do. Factor a
negative, or a positive? -(4x+14)

Iceycold12,

regarding 10x^2 + 35x - 4x - 14, first completely factor (over the usual integers)

the first pair to get 5x(2x + 7) - 4x - 14. If this trinomial can be factored, then

a (2x + 7) factor must be present in the factorization of the last pair of

terms. So, deliberately write (2x + 7) down as a factor of the last two terms

multiplied by some other yet-to-be-determined expression. Leave a couple of

spaces in front of the second (2x + 7) factor to allow for any addition or

subtraction signs, and any possible monomial factors:

5x(2x + 7) ____ (2x + 7).


Ask what does the first term, 2x, have to be multiplied by to equal -4x.

The answer is -2, so write that down as a subtraction sign followed by

a 2 right up against the (2x + 7) factor as:

5x(2x + 7) - 2(2x + 7).

Then, to make sure you have it correct up to that point,

ask if -2 multiplied by 7 gives -14. It does, so 5x(2x + 7) - 2(2x + 7)

is correct up to that point.
​

-------------------------------------------


Your other example:


\(\displaystyle 2k^2 + 3k - 16k - 24 \ = \)


\(\displaystyle k(2k + 3) - 16k - 24 \ =\)


\(\displaystyle k(2k + 3) \)____ \(\displaystyle (2k + 3) \ =\)


[What multiplied by 2k equals 16k? Answer: -8]


\(\displaystyle k(2k + 3) - 8(2k + 3)\)


[Check. Does -8 multiplied by 3 equal -24? Answer: Yes.]


Then k(2k + 3) - 8(2k + 3) is correct up to that point.

And then continue on to get (2k + 3)(k - 8).

 

[Iceycold12]

Lookagain, I appreciate that method it is easier to do. But my teacher wants the exact steps that he told me to do written, sucks I know but that's the way I need to understand it:

Ok think I got it now. -(4x-14) -> Transform it to -(4x+14) Factor both by 2 -> 2(2x+7) But since the negative is existant, I add the negative at the end to equal 5x(2x+7)-2(2x+7) -> (5x-2)(2x+7).

The reason I say I transform it to positive is because he says that in: 10x^2-35x-4x-14, to keep them as negative when I group them, -(4x+14) when distributed equals out to: -4x-14. I know, very confusing.​

By the way guys, why is the negative cut out? Why doesn't it remain, (10x^2-35x)+(-4x-14)? If you factor a -2 there you'd get -2(2x+7). Perfect it's done there. But I thought you could not factor a negative.

 

[Mrspi]

But I thought you could not factor a negative.

Of course you can remove a negative common factor!!

-8 - 4x = -4(2 + x)

-6y - 9z = -3(2y + 3z)

etc.

In fact, when you are doing this sort of factoring, you'll have MANY occasions where you will need to remove a common factor that is negative!

I'm guessing you might want to ask your teacher to explain his/her "take" on the process again, and listen very carefully to that explanation. I can't imagine a teacher saying that the negative is "cut out".....

 

[Iceycold12]

ok then problem solved

so when is it best to factor negatives, when there is two negative like in this problem? -4x-14

The rule with + and a - doesn't apply here right since it's in parenthesis: (10x^2+35x)+(-4x-14) edit: Answered my own question, the number in the end when factored results in a -

-2(2x+7)

All done now guys Happy to understand this now thanks for all the help. And lookagain, will use that method when I can.