# expanding binomials

• May 11th 2008, 01:52 PM
Nick87
expanding binomials
Promise no more questions today! ^_^

The reason I'm stuck on these is because I've completely forgotten what im doing, after returning to these from months ago.

I need to:

1) Express $x^2 - 8x + 15$ in the form $(x + a )^2 + b$ where "a" and "b" are the numbers to be found.

This is what I've got thus far:

$y = x^2 - 8x + 15$

$= (x - 4 )^2 - (4)^2 + 15 = 0$

$(x - 4)^2 - 16 + 15$

Then I need to plot for the original equation. Not alot (Thinking)

2) Expand the following using Pascal's Triangle

$(2 + x)^3$

I could manage the $(x+1)^n$ expansion but not sure how to go about this one.

Thanks
• May 11th 2008, 01:59 PM
Moo
Hello,

Quote:

Originally Posted by Nick87
Promise no more questions today! ^_^

The reason I'm stuck on these is because I've completely forgotten what im doing, after returning to these from months ago.

I need to:

1) Express $x^2 - 8x + 15$ in the form $(x + a )^2 + b$ where "a" and "b" are the numbers to be found.

This is what I've got thus far:

$y = x^2 - 8x + 15$

$= (x - 4 )^2 - (4)^2 + 15 = 0$

$(x - 4)^2 - 16 + 15$

Then I need to plot for the original equation. Not alot (Thinking)

Sorry if I misunderstood what you've said, but you got it :D

Quote:

2) Expand the following using Pascal's Triangle

$(2 + x)^3$

I could manage the $(x+1)^n$ expansion but not sure how to go about this one.

Thanks
So you know the formula for $(x+1)^n$

It's the same for $(2+x)^3$

$=C(3,0) 2^0 x^3+C(3,1)2^1 x^2+C(3,2) 2^2 x^1+C(3,3) 2^3 x^0$

Where $C(n,p)={p \choose n}$
• May 11th 2008, 02:07 PM
Nick87
So for the first one $a=4$ and $b=-16$?

It says to use my answer to sketch the graph of $y = x^2 - 8x + 15$ so what does that mean? Can't I just draw the graph like usual?
• May 11th 2008, 02:11 PM
Moo
Quote:

Originally Posted by Nick87
So for the first one $a=4$ and $b=-16$?

It says to use my answer to sketch the graph of $y = x^2 - 8x + 15$ so what does that mean? Can't I just draw the graph like usual?

Erm...

$y=(x-4)^2-16+15=(x-4)^2-1$

--> $a=-4$ and $b=-1$ (Wink)

To graph the thing, represent yourself :

- you know how to graph y=x²
- imagine that if you add a certain number to x, it's like translating to the right or the left the graph
- adding a constant (here, b) is like increasing or decreasing the curve, that is to say "go up or go down", because you add b to the ordinate of each point
• May 11th 2008, 02:18 PM
Nick87
Quote:

Originally Posted by Moo
Erm...

$y=(x-4)^2-16+15=(x-4)^2-1$

--> $a=-4$ and $b=-1$ (Wink)

*Slaps self* of course its -1!

So what I need to do now is plot the orignal graph but with the translation ${a \choose b}$?
• May 11th 2008, 02:23 PM
Moo
Quote:

Originally Posted by Nick87
*Slaps self* of course its -1!

So what I need to do now is plot the orignal graph but with the translation ${a \choose b}$?

Yep !
I have an hesitation, perhaps it's -a. Check on the graph which one it is :D
• May 11th 2008, 02:24 PM
Nick87
Thank you so much for all your help!