How do you foil $(n - 7) (3 n - 2)$ $(n-7)(3n-2)$ ?

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Answer 1 · 2015-08-26T03:57:38

$= 3 n^{2} - 23 n + 14$ $=color(blue)(3n^2-23n+14$

The expression is : $(n - 7) (3 n - 2)$ $(n−7)(3n−2)$

F O I L involves carrying out the below operations in the following order:

1) product of First terms : $(n and 3 n)$ $(color(blue)(n and 3n))$ +

2)product of Outside terms: $(n and - 2)$ $(color(blue)(n and -2))$ +

3)product of Inside terms: $(- 7 and 3 n)$ $(color(blue)(-7 and 3n))$ +

4)product of Last terms: $(- 7 and - 2)$ $(color(blue)(-7 and -2))$

1) $n \cdot 3 n = 3 n^{2} +$ $n *3n=color(blue)(3n^2) +$

2) $n \cdot - 2 = - 2 n +$ $n * -2=-color(blue)(2n)+$

3) $- 7 \cdot 3 n = - 21 n +$ $-7 * 3n = color(blue)(-21n)+$

4) $- 7 \cdot - 2 = 14$ $-7 *-2 = color(blue)14$

adding all the terms

$= 3 n^{2} - 2 n - 21 n + 14$ $=color(blue)(3n^2-2n-21n+14$
$= 3 n^{2} - 23 n + 14$ $=color(blue)(3n^2-23n+14$

How do you foil (n-7)(3n-2)(n−7)(3n−2)(n-7)(3n-2)?