# Confusing Question

• Jan 30th 2009, 03:39 PM
drguildo
Confusing Question
Can somebody please explain to me what is meant by the following question:

If $(8x-5)^5=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$, find $A+B+C+D+E+F$.

Thanks.
• Jan 30th 2009, 04:09 PM
Greenb
$(8x-5)^5=(8x-5)(8x-5)(8x-5)(8x-5)(8x-5)$

Multiply the parantheseses and you should get the constants (A,B,C....) infront of each x^n value where n=0,1,2,3,4,5.

Hope I helped you somewhat :)
• Jan 30th 2009, 04:23 PM
Chop Suey
You can use Binomial Theorem/Pascal's Triangle to expand. Look them up if you're not familiar with it.

To start you off:
$(8x+(-5))^5 = (8x)^5+5(8x)^4(-5)+10(8x)^3(-5)^2+10(8x)^2(-5)^3+5(8x)(-5)^4 +(-5)^5$
• Jan 30th 2009, 04:55 PM
drguildo
I'm more confused than when I posted the question. It's like I'm so confused that I don't even know what I'm confused about. (Headbang)

The question is from a book on basic algebra. The point the question is asked is before multiplying out parenthesis has been covered.

Letting $x = 1$ one gathers that $243 = 3^5 = (8(1) - 5)^5 = A+B+C+D+E+F$.

This doesn't make any sense to me. Why 1? Why not 20: $89466096875 = 155^5 = (8(20) - 5)^5 = A+B+C+D+E+F$

Should I learn from a different book? This is not the first time there's been a question like this.
• Jan 30th 2009, 05:07 PM
Chop Suey
Oh wow I can't believe I went about it methodically without thinking a bit more.

$(8x-5)^5=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$

This equality is supposed to hold for all x. If we input x = 1, we managed to make the right side into the expression in question. The left side would result in:
$(8(1)-5)^5 = (3)^5 = 243$

Nah, your book is fine so far. You did not need to multiply nor did you need Pascal's triangle *slaps forehead*