I need help understanding how to factor some expressions better.

[Vampire_Rex]

Here's the expression and steps to the solution: -4(x+16)^4+9(x+16)^2+x+16 = [-4(x+16)^3](x+16)+9(x+16)(x+16)+1(x+16) = [-4(x=16)^3+9(x+16)+1](x+16) = [-4(x+16)^3+9x+144+1](x+16) = [-4(x+16)^3+9x+145)(x+16)
I just need help understanding how to get from one step to the next.
I'd greatly appreciate any help whatsoever I can get.

 

[Vampire_Rex]

You do know that:
(x+16)^4 = (x+16)*(x+16)*(x+16)*(x+16) , yes?

Yes, I know that. Thank you for making more readable. The book is asking me under the title of the section "More Factoring" "An algebraic expression raised to different powers might appear in different terms. Factor out this expression raised to the lowest power." I've done some factoring so maybe a complete lesson isn't necessary. I just don't quite fully understand what I'm being asked to do here. here's the examples I was given in the book.

6(x+1)^2-5(x+1)

=[6(x+1)](x+1)-5(x+1)

=[6(x+1)-5](x+1)

=(6x+6-5)(x+1)

=(6x+1)(x+1)

Here's another:

2(x^2-6)^9+(x^2-4)^4+4(x^2-6)^3+(x^2-6)^2

=2(x^2-6)^7(x^2-6)^2+(x^2-6)^2(x^2-6)^2+4(x^2-6)(x^2-6)^2+1(x^2-6)^2

=[2(x^2-6)^7+(x^2-6)^2+4(x^2-6)+1(x^2-6)^2

=[2(x^2-6)^7+(x^2-6)^2+4(x^2-6)+1](x^2-6)^2

=[2(x^2-6)^7+(x^2-6)^2+4x^2-24+1](x^2-6)^2

=[2(x^2-6)^7+(x^2-6)^2+4x^2-23](x^2-6)^2

Maybe I'm just missing something basic and fundamental but I can't seem to get it for some reason. I've been pretty good at doing this on my own so far but I just need help with this. I don't know why I can't get it. (BTW I'm trying to learn algebra on my own before I actually start math in high school but I suppose I can't really do it completely alone especially if I'm using a regular book meant for self-teaching and not an actual textbook). Any help would be so so very deeply appreciated.

 

[Otis]

I just don't quite fully understand what I'm being asked to do here. here's [one of] the examples I was given in the book.

6(x+1)^2-5(x+1)

=[6(x+1)](x+1)-5(x+1)

=[6(x+1)-5](x+1)

=(6x+6-5)(x+1)

=(6x+1)(x+1)

Let's try Denis' suggestion, and make a substitution for the expression x+1, in the example above.

Let symbol u = x+1

Then, the original expression becomes

6u^2 - 5u

Factoring this gives

u(6u - 5)

That's easy to see, yes?

Now, reverse the earlier substitution by replacing symbol u with x+1, and we get the same result as the book.

(x+1)(6[x+1] - 5)

Apply the distributive property to 6[x+1]

(x+1)(6x + 6 - 5)

Simplify 6-5

(x+1)(6x + 1)

You can always substitute a single symbol for a lengthier expression; sometimes, doing this helps to recognize repeated factors more easily and to complete the algebraic steps. Reverse the substitution after factoring, to simplify.

With enough practice, you will be able to recognize these sorts of repeated factors (eg: binomials) and complete the factorization without making substitutions, but, until then, try substituting and see whether it helps!

 

[Vampire_Rex]

Thank you so much for your time and effort for helping me. I really do appreciate it a lot. Fortunately I was able to figure it out on my own after a whole day. But, thanks again, it means quite a bit to me.