How do you FOIL $(4 z - u) (3 z + 2 u)$ $(4z-u)(3z+2u)$ ?

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How do you FOIL $(4 z - u) (3 z + 2 u)$ $(4z-u)(3z+2u)$ ?

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Answer 1 · 2018-03-24T19:37:04

The FOIL'ed version of the factored expression is $12 z^{2} + 5 z u - 2 u^{2}$ $12z^2 +5zu-2u^2$

Explanation:

As you may know, FOIL stands for First Outer Inner Last. Using this Mnemonic, we'll work in order.

First:
$4 z \times 3 z = 4 \times 3 \times z \times z = 12 z^{2}$ $4zxx3z=4xx3xxzxxz=12z^2$

Outer:
$4 z \times 2 u = 4 \times 2 \times z \times u = 8 z u$ $4zxx2u=4xx2xxzxxu=8zu$

Inner:
$- u \times 3 z = - 1 \times 3 \times u \times z = - 3 u z = - 3 z u$ $-uxx3z=-1xx3xxuxxz=-3uz=-3zu$

Last:
$- u \times 2 u = - 1 \times 2 \times u \times u = - 2 u^{2}$ $-uxx2u=-1xx2xxuxxu=-2u^2$

Now, let's put them back together in the expression:

$12 z^{2} + 8 z u - 3 z u - 2 u^{2}$ $12z^2+8zu-3zu-2u^2$

Finally we combine like terms to get our finished expression:

$12 z^{2} + (8 - 3) z u - 2 u^{2} = 12 z^{2} + 5 z u - 2 u^{2}$ $12z^2+(8-3)zu-2u^2=color(red)(12z^2+5zu-2u^2)$

Answer 2 · 2018-03-24T19:40:07

$12 z^{2} + 5 z u - 2 u^{2}$ $12z^2 +5zu -2u^2$

Explanation:

The FOIL method:

$(4 z - u) (3 z + 2 u)$ $(4z−u)(3z+2u)$
$(4 z \cdot 3 z) + (4 z \cdot 2 u) + (- u \cdot 3 z) + (- u \cdot 2 u)$ $(4z*3z) +(4z * 2u) + (-u*3z) + (-u * 2u)$
$12 z^{2} + 8 z u - 3 z u - 2 u^{2}$ $12z^2 +8zu -3zu - 2u^2$
$12 z^{2} + 5 z u - 2 u^{2}$ $12z^2 +5zu -2u^2$

Hope this helps!

Answer 3 · 2018-03-24T19:59:15

$2$ $2$ $x^{2}$ $x^2$ + $5$ $5$ $u$ $u$ $z$ $z$ $-$ $-$ $2$ $2$ $u^{2}$ $u^2$

Explanation:

$F$ $F$ = First
$O$ $O$ = Outside
$I$ $I$ = Inside
$L$ $L$ = Last

$(4 z - u)$ $(4z-u)$ $(3 z + 2 u)$ $(3z+2u)$

Since $4 z and 3 z$ $4z and 3z$ are first you multiply them by each other...

$4 \cdot 3$ $4 * 3$ equals $12$ $12$
$z \cdot z$ $z * z$ equals $z^{2}$ $z^2$

$4 z and 2 u$ $4z and 2u$ are at the end so you would multiply them together
and get $8 u z$ $8uz$ it can also be $8 z u$ $8zu$ but later on you learn that the letter that comes first in the alphabet is usually at the beginning.

The inside would be
$- u and 3 z$ $-u and 3z$
In which $- u \cdot 3 z$ $-u * 3z$ would equal $- 3 u z$ $-3uz$

Finally, you solve the last two...
$- u \cdot 2 u$ $-u * 2u$
Which equals $- 2 u^{2}$ $-2u^2$

You're equation should now look like this;
$12 z^{2} + 8 u z - 3 u z - 2 u^{2}$ $12z^2+8uz-3uz-2u^2$

Subtract $8 u z and 3 u z$ $8uz and 3uz$ - They have common variables
You should get $5 u z$ $5uz$

Your final equation should look like this;
$12 z^{2} + 5 u z - 2 u^{2}$ $12z^2+5uz-2u^2$

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