# Algebraic Expression Factoring

• Aug 22nd 2008, 09:38 AM
cmf0106
Algebraic Expression Factoring
The book that I am working through does an absolutely pitiful job explaining what to do in these types of problems. I have no clue as to where to begin, if someone could please walk me through and explain why each step works it would be much appreciated, or if you have an easier method other than the book entirely.

$9(14x+5)^4+6x(14x+5)-(14x+5)=[9(14x+5)^3](14x+5)+6x(14x+5)-1(14x+5)$ $=[9(14x+5)^3+6x-1](14x+5)$

Other than this the only hint the book gives me is to "factor out each expression raised to the lowest power", but if that were the case wouldn't $(14x+5)^1$ be the lowest?
• Aug 22nd 2008, 10:15 AM
masters
Quote:

Originally Posted by cmf0106
The book that I am working through does an absolutely pitiful job explaining what to do in these types of problems. I have no clue as to where to begin, if someone could please walk me through and explain why each step works it would be much appreciated, or if you have an easier method other than the book entirely.

$9(14x+5)^4+6x(14x+5)-(14x+5)=[9(14x+5)^3](14x+5)+6x(14x+5)-1(14x+5)" alt="9(14x+5)^4+6x(14x+5)-(14x+5)=[9(14x+5)^3](14x+5)+6x(14x+5)-1(14x+5)" /> $=[9(14x+5)^3+6x-1](14x+5)$

Other than this the only hint the book gives me is to "factor out each expression raised to the lowest power", but if that were the case wouldn't $(14x+5)^1$ be the lowest?

You are correct. It would seem that the GCF (Greatest Common Factor) is $14x+5$
• Aug 22nd 2008, 10:18 AM
cmf0106
Well with that in mind how would I go about solving it?

$9(14x+5)^4+6x(14x+5)-(14x+5)$
• Aug 22nd 2008, 10:27 AM
Jhevon
Quote:

Originally Posted by cmf0106
Well with that in mind how would I go about solving it?

$9(14x+5)^4+6x(14x+5)-(14x+5)$

start by looking for what is common. what appears in all the terms....

lets look....ah! that thing in red

$9({\color{red}14x+5})^4+6x({\color{red}14x+5})-({\color{red}14x+5})$

now we want to factor it out. so we pull out the lowest power. i have 4 of those terms in the first term, and one in the second and third terms respectively. so i will pull out 1. if i take one from the first term, i am left with 3, so i put the 3rd power, if i take one from the second and third terms, i don't have any left, so i am just left with their respective coefficients. thus i get:

$\underbrace{(14x + 5)}_{\text{pull this out}}[9 \underbrace{(14x+5)^3}_{\text{left with 3 terms here}}+ \underbrace{6x - 1}_{\text{none left here}}]$
• Aug 22nd 2008, 10:32 AM
cmf0106
Is the book incorrect then when it lists $[9(14x+5)^3+6x-1](14x+5)$ as the answer then?

And thanks for your replies, I will look at them more closely later when its time to study again.
• Aug 22nd 2008, 10:34 AM
Jhevon
Quote:

Originally Posted by cmf0106
Is the book incorrect then when it lists $[9(14x+5)^3+6x-1](14x+5)$ as the answer then?

And thanks for your replies, I will look at them more closely later when its time to study again.

no, multiplication is commutative, so it doesn't matter which you write first. 2 times 3 is the same as 3 times 2, that sort of thing. they only wrote it the other way around. no problem
• Aug 22nd 2008, 11:42 AM
Moo
Quote:

Originally Posted by cmf0106
Other than this the only hint the book gives me is to "factor out each expression raised to the lowest power", but if that were the case wouldn't $(14x+5)^1$ be the lowest?

I'd say the book is correct too...

See the (14x+5) ? The greatest common factor contains (14x+5) raised to the lowest power appearing in the sum.
Among $(14x+5)^4$ and $(14x+5)$, the lowest power is 1. And that's what appears in the answer :

Quote:

$=[9(14x+5)^3+6x-1]{\color{red}(14x+5)}$
• Aug 22nd 2008, 11:48 AM
cmf0106
Honestly Im still very confused. Here lets take the example
$6(x+1)^2-5(x+1)$

To me it seems I should do something like this, first finding the GCF for all terms $(x+1)^1$

So factor out the $(x+1)^1$ from all terms and you would get
$6(x+1)-5$

However the book lists the correct answer as $(6x+1)(x+1)$ (Angry)

EDIT: it looks like for the final step in these type of problems, in this case $6(x+1)-5$ you distribute the 6 to get $(6x+6-5)=(6x+1)$ and then you just 'tac' on the GCF in this case (x+1) to get $(6x+1)(x+1)$ for your final answer. Is this the right approach?
• Aug 22nd 2008, 11:53 AM
Moo
Quote:

Originally Posted by cmf0106
Honestly Im still very confused. Here lets take the example
$6(x+1)^2-5(x+1)$

To me it seems I should do something like this, first finding the GCF for all terms $(x+1)^1$

So factor out the $(x+1)^1$ from all terms and you would get
$6(x+1)-5$

However the book lists the correct answer as $(6x+1)(x+1)$ (Angry)

Develop 6(x+1)-5, this gives 6x+1. :p
• Aug 22nd 2008, 11:58 AM
cmf0106
Well at any rate why do you not develop the problem found earlier?
$
[9(14x+5)^3+6x-1](14x+5)
$

IE raise each variable to the 3rd power, distribute the 9 combine like terms?
• Aug 22nd 2008, 12:05 PM
Moo
Quote:

Originally Posted by cmf0106
Well at any rate why do you not develop the problem found earlier?
$
[9(14x+5)^3+6x-1](14x+5)
$

IE raise each variable to the 3rd power, distribute the 9 combine like terms?

Because it's awful ^^

You can do it if you want, but it's not something you'd be asked...

I think it comes with practice that you know when to stop.
I'd say that if x is at the power 1, then you can develop without problem. Power 2, you can too, it's not too hard and it yields 3 terms which would simplify with some constant. Power 3 becomes difficult and quite unuseful because it will yield 4 terms... (Tongueout)
The thing is, I guess 2 terms in a factor (x+y, or x-y...etc) is reasonable. Because factoring is made for writing less terms in an expression.

Is it clear ? :(
• Aug 22nd 2008, 12:12 PM
cmf0106
Yah actually its all coming together now! Thank you everyone who contributed to this post and my learning. I just wish the author of this book could clarify the solutions like you guys (Happy).