factor by grouping

latresa31s

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Sep 4, 2005
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I just can't seem to understand this problem:

3xy +6y - 4x - 8

so here's how I started

3xy - 4x + 6y - 8 ( i group them to gether)

then I tried

3x(y - 4) + 6y - 8

what is wrong with this picture?
 

Daniel_Feldman

Full Member
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Sep 30, 2005
Messages
252
Hi,

I'm not really sure here. The problem with rewriting it the way you did

3xy +6y - 4x - 8

to

3xy - 4x + 6y - 8

is that the first term has an x and a y factor, where as the second and third terms have only an x and only a y factor, respectively. Thus, you did not really "group" them, but merely rewrote the expression. Supposing we looked at the expression the way you did:

3x(y - 4) + 6y - 8

is not correct. 3 is not a factor of four. This can easily be checked by multiplying 3x back into the first part of the expression:

3x(y-4)=3xy-12x, which is not equal to 3xy-4x.

I see two options for this problem:

One is to keep it as written:

3xy +6y - 4x - 8

We can now factor out a 3y for the first two terms (not the 3 is a factor of 6), which yields

3y(x+6)-4x-8

The other option is to rewrite it as you did:


3xy-4x+6y-8

Factorin out the x (and only the x!) from the first two terms yields

x(3y-4)+6y-8


These are the only two possibilities I see. I don't thin you can factor an entire expression because one term has xy, one has just x, one has just y, and one is a constant term.

Hope this helps,

-Daniel-
 

latresa31s

Junior Member
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135
Thanks it does
 

latresa31s

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can you help me

factor by grouping

15 + 5x - 3x - x^2

=5x - 3x + 15 - x^2

=x(5+3) + 5(3) (x)^2

= x(5+3) +5(3-x)
(I am not sure of this)

what step did I do wrong?
 

stapel

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For factoring "by grouping" to work, you have to note the useful groups.

. . . . .3xy +6y - 4x - 8

. . . . .3y(x + 2) - 4(x + 2)

. . . . .(x + 2)(3y - 4)

Eliz.

Edit: Correcting typo, noted below.
 

Daniel_Feldman

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Good call stapel, though you made a slight error....

3xy +6y - 4x - 8

=3y(x+2)-4(x+2)
=(3y-4)(x+2)


As for your other problem, latresa,

15 + 5x - 3x - x^2

Try making 5x-3x into one term, 2x.

Thus,

-x^2+2x+15

Factor out the negative sign

-(x^2-2x-15)

=-(x-5)(x+3)
 

latresa31s

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Thanks Daniel
 

latresa31s

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Messages
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Daniel,

ok, how would I factor out the neg.?

hum,

-x^2 + 2x +15

ok I got stuck and did it this way

5(3+x)- x(3-x)

(3 +x)(5 -x)

so could you please show me how to factor out the neg.
 

galactus

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A good way to factor this sort of problem is the following:

Let a=-1, b=2, c=15.

Find two numbers which when multiplied give 15 and which added give 2:

Ummm.....(3)(5)=15 and 5+(-3)=2

Now sub in for the middle (b) term:

-x^2+5x-3x+15

Group and factor out common terms:

-x(x-5)-3(x-5)=

(x-5)(-x-3)=

-(x-5)(x+3)
 

Denis

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Feb 17, 2004
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1,444
-x^2 + 2x +15
multiply by -1:
x^2 - 2x - 15 = 0
now factor:
(x - 5)(x + 3) = 0
x = 5 or x = -3
 

latresa31s

Junior Member
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Sep 4, 2005
Messages
135
Help Me Please

I have worked this problem and I and so confussed and fustrated. Help please

The problem is

4x^2 - 4x -24

this problem is to factor this problem completely.
 

Unco

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Jul 21, 2005
Messages
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Hi latresa31s,

4x^2 - 4x -24

Firstly, each term is a multiple of 4 so bring that out

4(x^2 - x - 6)

We now have a nice quadratic inside. Two numbers which sum to -1 and multiply to give -6 are -3 and +2.

So we have
4(x-3)(x+2)

This looks a little easier than what you have been doing in the above posts, perhaps a mental block? :D
 

latresa31s

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Sep 4, 2005
Messages
135
how did you get the 2.
 

Unco

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Jul 21, 2005
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As in the 2 from -3 and 2 sum to -1 and multiply to -6?

We need two numbers which mulitply to -6:
1 x -6
-1 x 6
2 x -3
3 x -2

We need them to add to -1.

3 + (-2) = -1 so 3 and -2 are our choices.

Does that help?
 

latresa31s

Junior Member
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Sep 4, 2005
Messages
135
yes it does thank you so much :lol:
 
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