# Is this considered factored form

1. Feb 17, 2016

### AllanW

1. The problem statement, all variables and given/known data
6(2x+5)^2(6x-1)^5+(2x+5)^3(30)(6x-1)^4
this is what i got from deriving (2x+5)^3(6x-1)^5 and now i have to express my answer in factored form, does this classify as factored from even though it has a '+' in it?

2. Relevant equations

3. The attempt at a solution

2. Feb 17, 2016

### RUber

You can take it one step further so that you only have products.

3. Feb 17, 2016

### SammyS

Staff Emeritus
By the way, the verb form for finding the derivative is "to differentiate".
Differentiation
The process of finding a derivative.​

That is not factored form.

There are common factors of: 6, (2x + 5), and (6x − 1) , to various powers.

4. Feb 17, 2016

### AllanW

okay, is there an easier way of combining the exponential binomials or do i have to expand all of them out completely and try to find a way to factor them back down?

5. Feb 17, 2016

### SammyS

Staff Emeritus
Expanding them would be counter productive.

Facotr out all of the common factors. Then see what's left and proceed from that point.

6. Feb 17, 2016

### Staff: Mentor

@AllanW, Problems involving differentiation should go in the Calculus & Beyond section. I have moved it for you.

7. Feb 17, 2016

### AllanW

i know, but my question was about the factoring part of the question which is why i had it under pre-calc

8. Feb 17, 2016

### RUber

As SammyS pointed out, you have something of the form $A^2B^5 + A^3B^4.$
Start by taking it down to $A^2B^4(B+A).$
Since B and A are both binomial terms, their sum will also be a binomial.

9. Feb 17, 2016

### AllanW

okay, so you're saying I should get:
(2x+5)^2(6x-1)^4((6x-1)+(2x+5))→(2x+5)^2(6x-1)^4(8x+4)
what do I do with my coefficients?

10. Feb 17, 2016

### RUber

That's right. Now do the same thing and keep your coefficients in there. SammyS pointed out that 6 was a common factor between them, so you will have to account for the leftover 5 by keeping with its binomial term.

11. Feb 17, 2016

### SammyS

Staff Emeritus
Don't ignore them!

12. Feb 17, 2016

### AllanW

so i should get 5(2x+5)^2(6x-1)^4(8x+4)?

13. Feb 17, 2016

### SammyS

Staff Emeritus
5 is not a common factor. 6 is a common factor

You have made some algebra errors.

14. Feb 17, 2016

### RUber

No.
You started with
$6A^2B^5 + 30A^3B^4.$
Which is $6A^2B^4(B+5A).$

15. Feb 17, 2016

### AllanW

okay, so it is apparent i have no idea how A^2B^5+A^3B^4=A^2B^4(B+A) perhaps we start there?

16. Feb 17, 2016

### RUber

In the form:
$A^2 B^5 + A^3 B^4,$
first look at term A.
There is a power of 2 and a power of 3. The smaller of these is 2, so $A^2$ is a common factor. Factor it out and you have:
$A^2( B^5 + A B^4).$
Next look at term B. Same rules apply, and you have a common factor in the sum of $B^4.$ Factor that out and you have:
$A^2B^4( B + A ).$
If you had coefficients, like 6 and 30, you would do the same thing...find the common factor, factor it out, and leave behind any part that was not factored.

If this is not making sense, please review the distributive property of multiplication...which tells you that:
a(b+c) = ab+ac.
This is what allows you to factor in this way...if you are given something like ab+ac, you can put it in factored form of a(b+c).

17. Feb 17, 2016

### AllanW

okay that makes sense, so would 48(2x+5)^2(6x-1)^4(2x+3) be my final answer?

18. Feb 17, 2016

### HallsofIvy

There is an "$A^2$" in the first term and "$A^3= A^2(A)$" in the second term so you can factor out $A^2$ from both, leaving "A" in the second: $A^2B^5+ A^3B^4= A^2(B^5+ AB^4). There is a "[itex]B^4$" in the second term and "$B^5= B^4(B)$" in the first term so you can factor out $B^4$ from both, leaving "B" in the first: $A^2(B^5+ AB^4)= A^3B^4(B+ A)$.

19. Feb 17, 2016

### RUber

That looks right to me.

20. Feb 17, 2016

### AllanW

Great, thanks!