Factor By Grouping

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mathdad

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Factor by grouping.

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

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

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

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

(x - 1)(x + 1)(x - 3)

Correct?
 

Romsek

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did you try multiplying out your result and seeing if it matched the original expression?
 

Jomo

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Factor by grouping.

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

x^2(x - 3) - (- x + 3)
x^2(x - 3) - (- x + 3) = x^2(x - 3) - 1(- x + 3) = x^3 -3x^2 +x -3 which is wrong
 

Jomo

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

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

mathdad

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

mathdad

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

Harry_the_cat

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Factor 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 from your previous line

(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?
see comments in red
 

lookagain

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Factor by grouping.

x^3 - 3x^2 - x + 3
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, there must be 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:

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

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

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

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

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

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Which is what I tried to get. Give me credit or walk away.
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.
 

mathdad

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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, there must be 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:

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

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

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

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

\(\displaystyle (x - 1)(x + 1)(x - 3)\)
Nicely done.
 

mathdad

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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.
Fair enough. I am also old school.
 

Otis

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... Give me credit or walk away.
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.

\(\;\)
 

mathdad

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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.

\(\;\)
I would not be surprised if I get banned from here, too. I am tired of having to explain every post.
 
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