Question

Factor
C^3-27

Answer 1

Are you familiar with the difference of cubes factoring rule?

Answer 2

c³-27=c³-3³

Answer 3

No, I am not.

Answer 4

If you have x^3 - y^3, then it factors to (x-y)(x^2 + xy + y^2)

Answer 5

Oh, okay. So how I set it up?

Answer 6

In your case (and from what josefelk wrote), we can see that x = C and y = 3

Answer 7

Plug those two in and simplify

Answer 8

But how do we know y=3?

Answer 9

because

C^3 - 27

becomes

C^3 - 3^3

Answer 10

compare

C^3 - 3^3

with

x^3 - y^3

Answer 11

you'll see that x = C and y = 3

Answer 12

Oh okay, thanks so much guys! :)

Answer 13

yw

Answer 14

Can you help me simplify it?

Answer 15

what do you get when you plug in x = C and y = 3

Answer 16

I get (C-3)(C^2+3c+3^2)

Answer 17

the only thing left to do is replace 3^2 with 9 (since 3^2 = 9) to get the final answer of (C-3)(C^2+3C+9)

Answer 18

Ohkay, I thought we had too go beyond that. Can you help me with one last problem?

Answer 19

sure

Answer 20

6x^7x+2

Answer 21

\[\Large 6x^{7x}+2\] ???

Answer 22

Sorry, 6x^2+7x+2

Answer 23

and what do you want to do here?

Answer 24

factor

Answer 25

@jim_thompson5910

Answer 26

multiply 6 and 2 to get 12

then find two numbers that multiply to 12 AND add to 7

Answer 27

Huh? Can you show it though equation?

Answer 28

you can also use the quadratic formula

Answer 29

which is

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 30

What? Sorry, I am so lost.

Answer 31

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(7)\pm\sqrt{(7)^2-4(6)(2)}}{2(6)}\]


\[\Large x = \frac{-7\pm\sqrt{49-(48)}}{12}\]


\[\Large x = \frac{-7\pm\sqrt{1}}{12}\]


\[\Large x = \frac{-7+\sqrt{1}}{12} \ \text{or} \ x = \frac{-7-\sqrt{1}}{12}\]


\[\Large x = \frac{-7+1}{12} \ \text{or} \ x = \frac{-7-1}{12}\]


Keep going to solve for x