Question

how to divide these help

Answer 1

http://prntscr.com/ls85dn

Answer 2

You should factor them first

Answer 3

individually?

Answer 4

There are two expressions. Factor both completely first.

Answer 5

1,-1

Answer 6

?

Answer 7

my bad (x-1)(x-1)

Answer 8

Can you post the steps to how you got that?

Answer 9

sorry i used a calc

Answer 10

i will look up an example and try to solve it give me a sec

Answer 11

for the second one is it (x-1)^2

Answer 12

correct

Answer 13

and the first one is the same but with a ^3?

Answer 14

(x-1)^3

Answer 15

What do you mean? Show all your steps

Answer 16

no thatsnot right

Answer 17

oh

Answer 18

hm

Answer 19

What does \(x^3 - 3x^2\) factor to?

Answer 20

x^2(x-3)

Answer 21

nvm

Answer 22

Hang on a minute let me post these there.

Answer 23

much better

Answer 24

Okay now we will deal with this right. \(x^3 - 3x^2 + 3x -1\)

Answer 25

correct

Answer 26

Give me a minute...

Answer 27

take your time

Answer 28

Let's see what happens if we do it this way:
\(x^3 - 1 - 3x^2 + 3x\)

Answer 29

If we re-write it this way then the first two terms is a difference of cubes since \(1^3 = 1\) so:
\(x^3 - 1^3 - 3x^2 + 3x\)

Answer 30

Remember that formula for difference of cubes is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

Answer 31

can we rewrite it this way for every factoring question ?

Answer 32

In other words, we can re-write the expression as:
\((x - 3)(x^2 - x + 1) - 3x^2 + 3x\)

Answer 33

No, there's different polynomials that require different approaches.

Answer 34

oh ok

Answer 35

Anyway, the next step is to factor the last two terms. Wanna give it a try? Don't use a calc though

Answer 36

Oh I made a mistake

Answer 37

The difference of cubes for the first two terms factor to:
\((x - 1)(x^2 - x + 1) - 3x^2 + 3x\)

Answer 38

There we go, now we can factor the last two terms.

Answer 39

how do i factor this (x−1)(x2−x+1)−3x2+3x? like how would i divide it?

Answer 40

Don't worry about dividing right now. We're factoring. I asked you to factor the last two terms of that expression. And please use \(\LaTeX\) to write. It's bad style to write the expressions the way you do without using caret symbols for the exponent.

Answer 41

ok

Answer 42

3x(x+1)

Answer 43

I can't even be mad with you because I made another mistake with my factorization, but with yours, you factored out $3x$ instead of $-3x$

Answer 44

lol

Answer 45

oh so 3x(-x+1)

Answer 46

Here's the correct factorization:
\(\begin{align*}
x^3 - 3x^3 + 3x - 1&=x^3 - 1^3 - 3x^3 + 3x \\
&=(x - 1)(x^2 + x + 1) - 3x(x - 1)\\
&=(x - 1)(x^2 + x + 1)
\end{align*}\)

Answer 47

You factor out \(-3x\) because its common to both terms

Answer 48

Now that we have the correct factors, NOW we can divide:
\(
\begin{align*}
\dfrac{x^3 - 3x^2 + 3x - 1}{x^2 - 2x + 1}&=\dfrac{(x - 1)(x^2 + x + 1)}{(x - 1)^2}
\end{align*}
\)

Answer 49

my eyes

Answer 50

And when you divide, you're mostly cancelling factors of one. Notice we can re-write what we factored as a product of one and the remaining factors
\(
\begin{align*}
\dfrac{(x-1)(x^2 + x + 1)}{(x - 1)^2} = \dfrac{x - 1}{x - 1} \cdot \dfrac{(x^2 + x + 1)}{x - 1}
\end{align*}
\)

Answer 51

And of course the factors of one cancel to get just:
\(\dfrac{x^2 + x + 1}{x - 1}\)

Answer 52

Which cannot be factored any further.

Answer 53

i cant cancel out variables?

Answer 54

You can only cancel out matching factors in the numerator and denominator but we canceled those already.

Answer 55

what about the two x's?

Answer 56

there is a positive one on the num and denom

Answer 57

you should really review your factoring. You should not be attempting to divide polynomials without mastering how to factor them.

Answer 58

yes will do thank you verry much !