Is this considered factored form

[AllanW]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[RUber]

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

 

[SammyS]

Homework Statement

6(2x+5)^2(6x-1)^5+(2x+5)^3(30)(6x-1)^4
this is what i got from deriving differentiating (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?

Homework Equations

The Attempt at a Solution

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.

 

[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?

 

[SammyS]

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?

Expanding them would be counter productive.

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

 

[Mark44]

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

 

[AllanW]

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

 

[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.

 

[AllanW]

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.

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?

 

[RUber]

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?

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.

 

[SammyS]

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?

Don't ignore them!

 

[AllanW]

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

 

[SammyS]

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

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

You have made some algebra errors.

 

[RUber]

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

 

[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?

 

[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).

 

[AllanW]

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

 

[HallsofIvy]

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?

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: [itex]A^2B^5+ A^3B^4= A^2(B^5+ AB^4).<br /> <br /> There is a "$B^4$" 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: $A^2(B^5+ AB^4)= A^3B^4(B+ A)$.[/itex]

 

[RUber]

That looks right to me.

 

[AllanW]

Great, thanks!

 

[epenguin]

This is just a special case. Convince yourself that whenever you differentiate uⁿv^m, you are always going to get something factorable. You could work out how you could differentiate in such a manner it doesn't split up and need to be recombined.