# Factors...and more Factors

• Oct 5th 2007, 10:51 AM
fluffy_penguin
Factors...and more Factors
I hate factors > <

1. http://i23.tinypic.com/15wflw1.jpg
Is that correct?

2. http://i24.tinypic.com/mmx2di.jpg
Is that correct? Do I have to take it a step further somehow?

3. http://i24.tinypic.com/106yko7.jpg
I have no clue what to do here.

4. http://i23.tinypic.com/2ymi1qa.jpg
Aside from C and D making no sense. The correct answer would be either A or B but it's B because it's the only one factored right? Isn't A like some GCF factor thing.

5. http://i20.tinypic.com/29y17jl.jpg
Is this correct?

Work:
(r^2-4tr) + (3wr -12tw)
gcf:r gcf:3w
r(r-4t) + 3w(r-4t)
woot
=(r-4t)(r+3w)

6. http://i20.tinypic.com/ztca6g.jpg
4+3xy-6y-2x

(4-2x) + (3xy-6y)
2(2-x) + 3y(x-2)

so now (2-x) and (x-2) are inverse or whatnot so multiplying one side by -1 would make the one on the right -3y(2-x) or would I leave the 3y alone?

Edit: got (2-x)(2-3y)
• Oct 5th 2007, 01:29 PM
Soroban
Hello, fluffy_penguin!

Ya done good!

Quote:

1) Factor: .\$\displaystyle 15x^3 + 5x\$

My answer: .\$\displaystyle 5x(3x^2+1)\$ . . . . Right!

2) Factor: .\$\displaystyle 7x^6y^4 + 21x^4y^3 + 14xy\$

My answer: .\$\displaystyle 7xy(x^5y^3+3x^3y^2 + 2)\$ . . . . Correct!

3) Factor out the greatest common factor. Simiplify the factors.
\$\displaystyle (4z-3)(z+4) - (4z-3)(z-3)\$

Note the common factor: .\$\displaystyle ({\color{red}4z-3})(z+4) - ({\color{red}4z-3})(z-3)\$

Factor it out: .\$\displaystyle (4z-3)\,\bigg[(z+4) - (z-3)\bigg]\$

Simplify: .\$\displaystyle (4z-3)\,\bigg[z + 4 - z + 3\bigg] \;=\;7(4z-3)\$

Quote:

4) Factor: .\$\displaystyle rt + rp + nt + np\$
Factor by grouping: .\$\displaystyle r(t+p) + n(t+p)\$

Then: .\$\displaystyle (t+p)(r+n)\$

Quote:

5. Factor:. \$\displaystyle r^2-1tw + 3wr - 4tr\$

Is this correct?
\$\displaystyle (r^2-4tr) + (3wr -12tw) \:=\:r(r-4t) + 3w(r-4t) \:=\:(r-4t)(r+3w)\$ . . . . Yes!

Quote:

6. Factor: .\$\displaystyle 4+3xy-6y-2x\$

My work:
\$\displaystyle (4-2x) + (3xy-6y) \:=\:2(2-x) + 3y(x-2)\$

so now (2-x) and (x-2) are inverse or whatnot,
then I would make the right -3y(2-x) . . . . Good!

Then you have: .\$\displaystyle 2(2 - x) - 3y(2 - x)\$

Factor out: .\$\displaystyle (2 - x)(2 - 3y)\$

Another approach . . .

We have: .\$\displaystyle 4 + 3xy - 6y - 2x\$

Rearrange terms: .\$\displaystyle 4 - 2x - 6y + 3xy\$

Factor by grouping: .\$\displaystyle 2(2 - x) - 3y(2 - x)\$

Therefore: .\$\displaystyle (2 - x)(2 - 3y)\$

• Oct 5th 2007, 01:46 PM
fluffy_penguin
Thanks for the help.

I was going to ask for #3 how you got positive 3 but I seemed to understand it now, the - had to be Multiplied
to both the z and the -3.

(4z-3)[(z+4)-(z-3)]
then
(4z-3)[z+4-z+3]