http://img165.imageshack.us/img165/8938/algos7.jpg

i get: x^2-2x-8=0

(x-4)(x+2)=0

?

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- Jan 13th 2008, 05:07 PMalgebra2algebra??
http://img165.imageshack.us/img165/8938/algos7.jpg

i get: x^2-2x-8=0

(x-4)(x+2)=0

? - Jan 13th 2008, 05:20 PMmr fantastic
You need to show your working - all of it. It's impossible to see the errors you've made and the misunderstandings you most likely have.

Here're some things to consider:

f(x + 2) = (x + 2)^2 - 2(x + 2) = expand and simplify.

f(2) = 0.

Therefore f(x + 2) - f(2) = .......

Therefore . - Jan 13th 2008, 05:27 PMalgebra2
first i did:

x^2-2x(x+2) - 2x^2-4x / x

second i did:

x^3 + 2x^2- 2x^2 - 4x - 2x^2 - 4x

third i simplified and got:

x^3 - 2x^2 - 8x / x

then i divided it all by x and got:

x^2 - 2x - 8

then i set it up with 0:

x^2 - 2x - 8 = 0

finally i put it into this form:

(x-4)(x+2) = 0

sorry, but i am bewildered with what you got and how you got it. could you please show step by step how you got that? - Jan 13th 2008, 05:43 PMmr fantastic
Oh, I see. Lucky you replied.

I now know exactly what your misunderstanding is. It's a bad one. But maybe simple to fix. Read the following very carefully.

f(x+2) does NOT mean f times (x + 2), that is, f(x + 2) does NOT mean (x^2 - 2x) (x + 2) .......

If f(x) = x^2 - 2x, then I hope we can agree that f(a) = a^2 - 2a, f(b) = b^2 - 2b etc. f(apple) = (apple)^2 - 2(apple) etc. Do you see how it works? You replace the x in the rule with whatever is in ().

f(x + 2) = (x + 2)^2 - 2(x + 2). I have replaced x in the rule with (x + 2).

Then f(x + 2) = x^2 + 4x + 4 - 2x - 4 = x^2 + 2x.

Similarly, f(2) does not mean f times 2. f(2) = 2^2 - 2(2). I have replaced x in the rule with 2.

It is essential that you revise functional notation and what it means or you are in for a world of pain ..... - Jan 13th 2008, 06:08 PMalgebra2
oh now i see.

so its: (x+2)(x+2) - 2(x+2) / x

= x^2 + 4x + 4 - 2x - 4 - 2^2 - 4 / 2

= x^2 + 2x / x

divide the x and i get "x+2"

thanks for clearing things up.