know I am supposed to use the FOIL method.But is this right? I am really not sure I did this right, something just seems off about it.
So i would Multiply
x/3 * 3x/4=x^2/4
x/3 *-3/5= -x/5
3/4 *3x/4=9x/4
3/4 * -3/5
I know to add both of my bottom numbers have to be the same so the common factors of them would be 20
so I get x^2-4x/20+45x/20-9/20=
x^2+41x/20-9/20
How do you Multiply this?
(x/3+3/4)(3x/4-3/5)
4 answers
know I am supposed to use the FOIL method.But is this right? I am really not sure I did this right, something just seems off about it.
So i would Multiply
x/3 * 3x/4=x^2/4
x/3 *-3/5= -x/5
3/4 *3x/4=9x/4 wouldn't this be 9x/16?
3/4 * -3/5
I know to add both of my bottom numbers have to be the same so the common factors of them would be 20
so I get x^2-4x/20+45x/20-9/20=
x^2+41x/20-9/20
So i would Multiply
x/3 * 3x/4=x^2/4
x/3 *-3/5= -x/5
3/4 *3x/4=9x/4 wouldn't this be 9x/16?
3/4 * -3/5
I know to add both of my bottom numbers have to be the same so the common factors of them would be 20
so I get x^2-4x/20+45x/20-9/20=
x^2+41x/20-9/20
Ohh, I knew something looked off...so it should be x^2-5x/80-9/20
What happened to the 4 (x^2/4) from the first term? You started with that but I think your dropped it.
For the middle term, it is -x/5 + 9x/16 and that part becomes (-16x + 45x)/80 which is 29x/80.
The last term is 9/20.
Then you can get rid of the 4, 80, and 20 as denominators by multiplying through by 80.
Check my thinking. Check my arithmetic.
For the middle term, it is -x/5 + 9x/16 and that part becomes (-16x + 45x)/80 which is 29x/80.
The last term is 9/20.
Then you can get rid of the 4, 80, and 20 as denominators by multiplying through by 80.
Check my thinking. Check my arithmetic.
1 answer