Question

(14x^3 - 21x^2 + 28x) / (7x)
Help me?

Answer 1

@eliassaab

Answer 2

@mossyfish

Answer 3

Hint:\[14 x^3-21 x^2+28 x=7 x \left(2 x^2-3 x+4\right) \]

Answer 4

So, let's start off by dividing everything by 7. Forget about the variables for now.

14/7=?
21/7=?
28/7=?

Answer 5

is it 21x^4 - 28x^3 + 35x^2??

Answer 6

We're dividing :)

Answer 7

Well, 14/7 = 7
21/7 = 3
28/7 =4

Answer 8

14/7 is not 7. We can check this by multiplying 7 by 7, we get 49.
You're right that 21/7=3 and that 28/7 is 4 :)

Answer 9

Sorry, it's 2. I was thinking subtraction.

Answer 10

That's alright!!
So, now that we've divided the 7 part, let's divide by x :)

What is
2x^3/x
3x^2/x
4x/x ?

Answer 11

I don't know how to do that...

Answer 12

Perfectly okay :) When we divide a variable by the same variable, we just take it away.

I'll do the first one as an example.

2x^3 means we have x*x*x. If we divide by 1 x, we now have x*x. It's 2x^2 Does this make sense? This website goes a little more in depth about dividing variables
http://www.dummies.com/how-to/content/dividing-variables-in-algebra.html

Answer 13

So the second one would be 3x^1?

Answer 14

Exactly! Except, we don't need to have that ^1 there because it's indicating that there's only one x... So it's just 3x :)

Answer 15

So the last one would be just 4x too?

Answer 16

The last one has one x to start with, so it'd be just 4 :)

Answer 17

So, great, we have all of our numbers divided! Let's plug this back into the equation.

\[2x^2-3x+4\]

Answer 18

Would that be it? or do we have to solve it from there?

Answer 19

Unless if we have a value for x, we can't really solve any further. So, that'd be it!! Great job