
Urgent Maths Homework :(
Hey,
I aint to good at maths and need some help on basic algebra lol.
Can someone explain to me the method of Removing the brakets and simplifying this:
(a  7)(a +5)
The answer is a2  2a  35 (a2 is a squared)
Please Help me understand how you get that :(
Thanks alot
TNT

(a + b)(c + d) = ac + ad + bc + bd
At Dutch schools we call this method: Vogelbekje. In English it is something like "mouth of a bird". If you draw lines from a to c, a to d and then b to c and b to d. The first two lines above the formula and the last two under the formula, you will notice why.

(Term1 + Term2) (Term3 + Term4)
=
(Term1 * Term3) + (Term1 * Term4) + (Term2 * Term3) + (Term2 * Term4)
Obviously, some of the "+" are "" depending on the signs in your original equation.
In your case...
T1 * T2 is (a * a) i.e. a^2
T1 * T3 is (a * 5)
T2 * T3 is (7 * a)
T2 * T4 is (7 * 5) i.e. 35
(a * 5) + (7 * a) simplifies to 2a
So a^2  2a 35
*** EDIT ***
Don't you just hate it when someone is typing their answer at the same time as you!!!

the quantity of a7 times the quantity of a+5
When you have 'quantity of' with multiplication and division, you have to make sure that you perform all of the multiplication and division before you do any addtion and subtraction (common in all mathematics).
In your problem...
(a7)(a+5)
you start with the first digit (or representative thereof) which is +a in this case. You take this and multiply it by the other side...
(a)(a+5)
to get
a^2 + 5a ... (or a squared plus five a )
then you do the same with the negative seven...
(7)(a+5)
to get
7a  35 ... (or negative seven a plus a negative 35)
put them all together
a^2 + 5a  7a  35 (a squared plus five a plus a negative seven a plus a negative 35
combine the like terms which are 5a and 7a to get 2a
and you are left with
a^2  2a  35
does that help any?
edit... yeah I hate it... especially when you type more and two people beat you to it :) :D

Ok,
Thanks alot guys
Cheers
TNT

I was always taught to use FOIL for multiplying 2 binomials(sp?)
you multipy the F(irst) part of each parenthese then the outer most 2, followed by the inner most two, and finally the second one in each parenthese.
ie...
(1+2)*(3+4)=1*3 + 1*4 + 2*3 + 2*4