(x+y)^4
My brain is going to explode
Ok well, we can do it your way first.
Just continue foiling you're almost there.
(x^2 + xy + xy+ y ^2) (x^2 + xy + xy+ y ^2)
You get
(x^2 + 2xy + y^2) (x^2 + 2xy + y^2)
Just finish it up.
(x^2)(x^2)+(x^2)(2xy)+(x^2)(y^2)..........
It's long, I know but if you forget pascal's triangle during a test, this is your best hope, write neatly and legibly.
As for Pascal's triangle...
You know that it's
1
11
121
1331
And so on...
If you think about it, these are the powers of 11.
11^0 = 1
11^1 = 11
11^2 = 121
.....
That's how I remember which level I take the numbers from.
Since you're expanding to 4.
You take it from the 4th level which is 1331
I recommend you write it out this way
1 * x * y
3 * x * y
3 * x * y
1 * x * y
Now you know from (x+y)^2, you get x^2 + 2xy + y^2
You would've gotten it this way.
1 * x^2 * y^0 = 1(x^2)(y^0) = x^2
2 * x^1 * y^1 = 2(x^1)(y^1) = 2xy
1 * x^0 * y^2 = 1(x^0)(y^2) = y^2
[Then you add them all up.
x^2 + 2 xy + y^2, that's the answer from (x+y)^2]
Realize how x starts from 2 and ends at 0 whereas y starts from 0 and ends at 2.
So does the same thing for this.
1 * x * y
3 * x * y
3 * x * y
1 * x * y
X starts from 4 and y starts from 0.
I hope this helps.
Giving just the final answer doesn't really help anyone...
If you know the Binomial Theorem, then you should know that
$\displaystyle (a + b)^n = \sum_{r = 0}^{n}{\left(^n_r\right)a^{n - r}b^r}$.
So in this case, $\displaystyle a = x, b = y, n = 4$.
So you have $\displaystyle (x + y)^4 = \sum_{r = 0}^{4}{\left(^4_r\right)x^{4 - r}y^r}$
$\displaystyle = \left(^4_0\right)x^4y^0 + \left(^4_1\right)x^3y^1 + \left(^4_2\right)x^2y^2 + \left(^4_3\right)x^1y^3 + \left(^4_4\right)x^0y^4$
$\displaystyle = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$.