FOIL question can you tell me step by step how to FOIL (x+3)(x+3)(x+3)?

Answer 1

Well you can't explicitly use FOIL for all of those terms. You can use FOIL for the first two groups to get:

(x*x) + (x*3) + (3*x) + (3^2) which simplifies to:
(x^2 + 6*x + 9)

And then you still have one more (x+3) to multiply in, which gives you:
(x^2 + 6*x + 9)*(x+3)

To solve this one you need to multiply the x times all of the terms in the first part and then add that to 3 times all of the terms in the first part:
[(x^2 + 6*x + 9)*x] + [(x^2 + 6*x + 9)*3]
x^3 + 6x^2 + 9x + 3x^2 + 18x + 27 Simplifies to:
x^3 + 9x^2 + 27x + 27

Answer 2

You can not FOIL 3 sets of parentheses, only 2. Here is how you would do (x+3)(x+3)......

(x+3)(x+3)
First ~ x • x = x squared
Outer ~ x • 3 = 3x
Inner ~ 3 • x = 3x
Last ~ 3 • 3 = 9
(x+3)(x+3) = x squared + 6x + 9

Answer 3

(x + 3) (x ² + 3 x + 3 x + 9)
(x + 3) (x ² + 6 x + 9)
x ³ + 6 x ² + 9 x + 3 x ² + 18 x + 27
x ³ + 9 x ² + 27 x + 27

Answer 4

start first with the first two (x+3)(x+3) :

first : x * x = x^2
outer: x * 3 = 3x
inner: 3 * x = 3x
last: 3 * 3 = 9

then your equation is = x^2 + 3x + 3x + 9
= x^2 + 6x + 9

now for the finishing touch: (x^2 + 6x + 9)(x+3)
same concept as the first
1: x^2 * x = x^3
2: x^2 * 3 = 3x^2
3: 6x * x = 6x^2
4: 6x * 3 = 18x
5: 9 * x = 9x
6: 9 * 3 = 27

add them all up: x^3 + 3x^2 + 6x^2 + 18x + 9x + 27
= x^3 + 9x^2 + 27x + 27

hope that helps.

Answer 5

(x+3)(x+3)(x+3) =

[ x(x+3) + 3(x+3) ](x+3) =

[ (x*x + 3*x) + (3*x + 3*3) ](x+3) =

[ x*x + 6*x + 9 ](x+3) =

(x+3)[ x*x + 6*x + 9 ] =

(x*x*x + 6*x*x + 9*x) + (3*x*x + 3*6*x + 3*9) =

(x^3 + 6x^2 + 9x) + (3x^2 + 18x + 27) =

x^3 + (6x^2 + 3x^2) + (9x + 18x) + 27 =

x^3 + 9x^2 + 27x + 27

Answer 6

break it down into two components:
First Component: (x+3)(x+3)

now FOIL that first; once you get the answer to that

then multiply the answer by the remaining (x+3)

Answer 7

the answers x^3 + 6x^2 +3x + 27

First you multiply the X across to all the other x's to get x^3 then you multiply it to the +3 part of the second eq, then do it again to 6X but because you did it two times it becomes 6x^2 then you mutiply it to three to make it 3x then you multiply three across to make it 27.

Answer 8

(x+3)(x+3)(x+3)=
first do (x+3)(x+3)
first outer inner last= x^2+3x+3x+9
=(x^2+6x+9)(x+3)= x^3+6x^2+9x
+ 3x^2+18x+27=


x^3+9x^2+27x+27

Answer 9

you would multiply x with the other x and then again w/ the 3rd x, and do that but also multiply w/ the 3rd parenthesis, u get it? it would be a very long equation but u can simplify after