How do I factorise this? (1 Viewer)

[Hypem]

I've never really learnt how to do this. A step-by-step process would be nice

http://puu.sh/2kkFm

Thanks

 

[Hypem]

The back of the book says it becomes:

http://puu.sh/2kkPs

They also had the 168e term before the 28e term before it was factorised, if that helps with anything.

 

[Praer]

<a href="http://www.codecogs.com/eqnedit.php?latex=28 e^2^x (e^2^x @plus; 1)^5 * (e^2^x @plus;1 @plus; 6(e^2^x))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" title="28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" /></a>

 

[Hypem]

<a href="http://www.codecogs.com/eqnedit.php?latex=28 e^2^x (e^2^x @plus; 1)^5 * (e^2^x @plus;1 @plus; 6(e^2^x))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" title="28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" /></a>

...but what did you do to get that answer?

 

[Praer]

Ok
1. Factor out the <a href="http://www.codecogs.com/eqnedit.php?latex=28e^2^x(e^2^x@plus;1)^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?28e^2^x(e^2^x+1)^5" title="28e^2^x(e^2^x+1)^5" /></a>
2. you're left with <a href="http://www.codecogs.com/eqnedit.php?latex=e^2^x@plus;1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?e^2^x+1" title="e^2^x+1" /></a> and <a href="http://www.codecogs.com/eqnedit.php?latex=6(e^2^x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?6(e^2^x)" title="6(e^2^x)" /></a>
3. add the left ones, and times it to <a href="http://www.codecogs.com/eqnedit.php?latex=28e^2^x(e^2^x@plus;1)^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?28e^2^x(e^2^x+1)^5" title="28e^2^x(e^2^x+1)^5" /></a>
4. and you're left with <a href="http://www.codecogs.com/eqnedit.php?latex=28 e^2^x (e^2^x @plus; 1)^5 * (e^2^x @plus;1 @plus; 6(e^2^x))" target="_blank"><img src="http://latex.codecogs.com/gif.latex?28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" title="28 e^2^x (e^2^x + 1)^5 * (e^2^x +1 + 6(e^2^x))" /></a>

Done!

 

[Hypem]

Ohh, I see what's happening now.

Thanks

 

[Praer]

Btw, you're hsc is this coming year...
This is something we learnt in year 10 -.-"

 

[Praer]

No problem, and you better work hard!

 

[Hypem]

Btw, you're hsc is this coming year...
This is something we learnt in year 10 -.-"

I know how to factorise most things, I think this was just a little more complex than the others.

 

[Hypem]

Also, you left the 6(e^2x) term and the e^2x term separate, they should just become 7e^2x, right?

 

[jarrodoliver1]

What the fark? This is 2U? What topic are we expected to learn this in?

 

[Hypem]

What the fark? This is 2U? What topic are we expected to learn this in?

Yep, in the Maths in Focus textbook. Exercise 4.2, Question 2.

It's not some sort of factorisation exercise, it's just part of the Exponentials and Logarithms chapter where you have to find the second derivative of http://puu.sh/2klat

 

[jarrodoliver1]

Yep, in the Maths in Focus textbook. Exercise 4.2, Question 2.

It's not some sort of factorisation exercise, it's just part of the Exponentials and Logarithms chapter where you have to find the second derivative of http://puu.sh/2klat

Oh that's okay then, we start logs after the holidays