- Thread starter mathdad
- Start date

- Status
- Not open for further replies.

- Joined
- Dec 30, 2014

- Messages
- 3,698

x^2(x - 3) - (- x + 3) = x^2(x - 3) - 1(- x + 3) = x^3 -3x^2 +x -3 which is wrongFactor by grouping.

x^3 - 3x^2 - x + 3

x^2(x - 3) - (- x + 3)

- Joined
- Dec 30, 2014

- Messages
- 3,698

x^2(x - 3) + (x - 3) = x^2(x - 3) + 1(x - 3) = (x^2 + 1)(x-3) which is not what you got!x^2(x - 3) + (x - 3)

(x^2 - 1)(x - 3)

- Joined
- Apr 24, 2015

- Messages
- 716

No. I did not multiply the result.did you try multiplying out your result and seeing if it matched the original expression?

- Joined
- Apr 24, 2015

- Messages
- 716

Which is what I tried to get. Give me credit or walk away.x^2(x - 3) + (x - 3) = x^2(x - 3) + 1(x - 3) = (x^2 + 1)(x-3) which is not what you got!

- Joined
- Mar 16, 2016

- Messages
- 1,431

see comments in redFactor by grouping.

x^3 - 3x^2 - x + 3

x^2(x - 3) - (- x + 3) this should be x^2(x - 3) + (-x + 3)

x^2(x - 3) + (x - 3) ……………………. x^2(x - 3) + -1(x - 3)

(x^2 - 1)(x - 3) ………………………. this is correct, but doesn't follow fromprevious lineyour

(x - 1)(x + 1)(x - 3) ……………………….. yes this is correct, but you have made several mistakes in your working.

Check your factorising by expanding this out and see if you get the original expression.

Correct?

Address the potential confusion head-on for the sign of the quantity that you are factoring out in the second pair.Factor by grouping.

x^3 - 3x^2 - x + 3

Then, there should be no doubts:

\(\displaystyle x^3 - 3x^2 - x + 3 \ = \)

\(\displaystyle x^2(x - 3) - x + 3 \)

For this expression to factor, there

with a few spaces in front of it for a plus sign or subtraction sign (depending), and a constant for a placeholder.

\(\displaystyle x^2(x - 3) \ \ \ \ \ (x - 3) \ \)

What multiplied by x equals -x? Negative one does. Write this in as a subtraction sign and 1 next to the second

(x - 3) factor.

\(\displaystyle x^2(x - 3) - 1(x - 3) \ \)

Lastly, check to see if (-1) multiplied by the (-3) of (x - 3) gives 3, the last term of the original polynomial expression.

It does.

Factor out the common factor of (x - 3), either to the right, or to the left, of \(\displaystyle \ (x^2 - 1)\).

Then, \(\displaystyle (x^2 - 1)(x - 3) \ = \)

\(\displaystyle (x - 1)(x + 1)(x - 3) \)

It just took me a lot of words and some minutes of hen-and-peck typing to explain this, but a newer student could

casually write out the steps on paper to fully factoring this by the grouping method in about 20 seconds, give or

take some seconds.

- - - - - - - - - - - - - - - - - - - - - --

This could also been factored by grouping this way, with keeping the highest degree term first:

\(\displaystyle x^3 - 3x^3 - x + 3 \ = \)

\(\displaystyle x^3 - x - 3x^2 + 3 \ = \)

\(\displaystyle x(x^2 - 1) \ - \ 3(x^2 - 1) \ =\)

\(\displaystyle (x^2 - 1)(x - 3) \ = \)

\(\displaystyle (x - 1)(x + 1)(x - 3)\)

Last edited:

- Joined
- Dec 30, 2014

- Messages
- 3,698

No, I will not give you credit if you are wrong. I hate when teachers do that. When you are wrong I will tell that you are wrong and when you are right I will say that.Which is what I tried to get. Give me credit or walk away.

- Joined
- Apr 24, 2015

- Messages
- 716

Thank you for letting me know.see comments in red

- Joined
- Apr 24, 2015

- Messages
- 716

Nicely done.Address the potential confusion head-on for the sign of the quantity that you are factoring out in the second pair.

Then, there should be no doubts:

\(\displaystyle x^3 - 3x^2 - x + 3 \ = \)

\(\displaystyle x^2(x - 3) - x + 3 \)

For this expression to factor, theremustbe an (x - 3) factor in the second pair, so deliberately write it there,

with a few spaces in front of it for a plus sign or subtraction sign (depending), and a constant for a placeholder.

\(\displaystyle x^2(x - 3) \ \ \ \ \ (x - 3) \ \)

What multiplied by x equals -x? Negative one does. Write this in as a subtraction sign and 1 next to the second

(x - 3) factor.

\(\displaystyle x^2(x - 3) - 1(x - 3) \ \)

Lastly, check to see if (-1) multiplied by the (-3) of (x - 3) gives 3, the last term of the original polynomial expression.

It does.

Factor out the common factor of (x - 3), either to the right, or to the left, of \(\displaystyle \ (x^2 - 1)\).

Then, \(\displaystyle (x^2 - 1)(x - 3) \ = \)

\(\displaystyle (x - 1)(x + 1)(x - 3) \)

It just took me a lot of words and some minutes of hen-and-peck typing to explain this, but a newer student could

casually write out the steps on paper to fully factoring this by the grouping method in about 20 seconds, give or

take some seconds.

- - - - - - - - - - - - - - - - - - - - - --

This could also been factored by grouping this way, with keeping the highest degree term first:

\(\displaystyle x^3 - 3x^3 - x + 3 \ = \)

\(\displaystyle x^3 - x - 3x^2 + 3 \ = \)

\(\displaystyle x(x^2 - 1) \ - \ 3(x^2 - 1) \ =\)

\(\displaystyle (x^2 - 1)(x - 3) \ = \)

\(\displaystyle (x - 1)(x + 1)(x - 3)\)

- Joined
- Apr 24, 2015

- Messages
- 716

Fair enough. I am also old school.No, I will not give you credit if you are wrong. I hate when teachers do that. When you are wrong I will tell that you are wrong and when you are right I will say that.

- Joined
- Apr 22, 2015

- Messages
- 1,772

That's not very nice. It's an example of one of the same types of bad behavior that got you banned from two other math sites.... Give me credit or walk away.

\(\;\)

- Joined
- Apr 24, 2015

- Messages
- 716

I would not be surprised if I get banned from here, too. I am tired of having to explain every post.That's not very nice. It's an example of one of the same types of bad behavior that got you banned from two other math sites.

\(\;\)

- Status
- Not open for further replies.