Multiply by FOIL method?

Question

[(1/2)x + (1/3)] [(1/2)x - (1/4)]

Answer 1

Multiply first terms: (1/2)x * (1/2)x = (1/4)x^2
Multiply outer terms: (1/2)x * -(1/4) = -(1/8)x
Multiply inner terms: (1/3) * (1/2)x = (1/6)x
Multiply last terms: (1/3) * -(1/4) = - (1/12)
(1/4)x^2 - (1/8)x + (1/6)x - 1/12
(1/4)x^2 - (3/24)x + (4/24)x - 1/12
Final answer: (1/4)x^2 + (1/24)x - 1/12

Answer 2

I wish people would quit teaching FOIL. It's a trick. It's a crutch. And its use is limited to binomials.

Remember the distributive property of multiplication over addition. That is. a(b+c)=ab+ac.

Thus
[(1/2)x + (1/3)] [(1/2)x - (1/4)]
becomes
[(1/2)x + (1/3)] [(1/2)x] + [(1/2)x + (1/3)] [(-1/4)]

And applying it again.
[(1/2)x(1/2)x + (1/3)(1/2)x] + [(1/2)x(-1/4) + (1/3)(-1/4)]

Now, doing a little arithmetic,
(1/4)x² + (1/6)x + (-1/8)x + -1/2

Doing that distributive thing again
(1/4)x² + (1/6 + -1/8)x + -1/2

More arithmetic
(1/4)x² + (1/24)x + -1/2

You might write that
x²/4 + x/24 + -1/2
or
(6x²+x-12)/24
But you know what your teacher is looking for.

Answer 3

First:
(1/2x)(1/2x)
1/4xˆ2

Outside:
(1/2x)(-1/4)
-1/8x

Inside:
(1/3)(1/2x)
1/6x

Last:
(1/3)(-1/4)
-1/12

Add all of the 4 terms together:
(1/4xˆ2) + (-1/8x) + (1/6x) + (-1/12)
1/4xˆ2 + 1/24x - 1/12
Multiply by 24 to get rid of the denominators:
6xˆ2 + x - 2
Plug the values 6 for a, 1 for b, and -2 for c, into the quadratic formula to solve for x:
(-b±√bˆ2-4ac) ÷ (2a)
(-1±√1+48) ÷ (12)
(-1 ± 7) ÷ 12
{1/2, -2/3}

Answer 4

2
=x -1/8 x +1/6 -1/12
2
=x +1/2x- 1/12

Answer 5

= (1/4).x² - (1/8).x + (1/6).x - (1/12)
= (1/4).x² + (1/24).x - (1/12)
= (1/24).(6x² + x - 2)
= (1/24).(3x + 2).(2x - 1)

Answer 6

sorry sorry mis read the qn, wait lemme do again

uhh nvm since linduh has done it then nvm

Answer 7

front, outside, inside , last