Hey, I have a question, you can factor using guess and check or grouping, here it is

-13a + 12 - 4a^2

the answer is -(a+4)(4a-3)

Not sure how to get it using either one, please help! Thanks!

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- Jan 13th 2008, 09:20 PMMathMackFactoring polynomials
Hey, I have a question, you can factor using guess and check or grouping, here it is

-13a + 12 - 4a^2

the answer is -(a+4)(4a-3)

Not sure how to get it using either one, please help! Thanks! - Jan 13th 2008, 10:02 PMjanvdl
First re-arrange:

$\displaystyle -4a^2 - 13a + 12 $

Now factor out the negative sign

$\displaystyle -(4a^2 + 13a - 12) $

Now we should take the $\displaystyle 4a^2 $ term and factorise that, keeping in mind that it is not yet complete. (The $\displaystyle x_{1} $ and $\displaystyle x_{2} $ terms represent the constants)

$\displaystyle -(4a + x_{1})(a + x_{2}) $

Now we should see which constants give us a product of 12.

(1 x 12 ; 2 x 6 ; 3 x 4)

Now from those options we should find which also satisfies the middle term.

Can you take it from here? Just do it step by step. - Jan 13th 2008, 10:03 PMMathMack
I also have 4 more, any help is appreciated!

2.) n^3-n^2+n-1

3.) 4a(a+2) - 8 (a+2)

4.) -6x^3 +10x^2 - 4x^4

5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle. Thanks again - Jan 13th 2008, 10:08 PMMathMack
kinda still stuck on that first one. Thanks for your help

- Jan 13th 2008, 10:11 PMjanvdl
2)$\displaystyle n^2(n - 1) + 1(n - 1)$

$\displaystyle = (n^2 + 1)(n - 1)$

3) $\displaystyle (4a - 8)(a + 2)$

4) $\displaystyle -2(3x^3 - 5x^2 + 2x^4)$

$\displaystyle = -2x^2(3x - 5 + 2x^2)$

$\displaystyle = -2x(2x^2 + 3x - 5)$

$\displaystyle = -2x(2x + 5)(x - 1)$

5) $\displaystyle 6x^2 + x - 1$

We know that this value is equal to 2 values multiplied by each other. So let's factorise it:

$\displaystyle (3x - 1)(2x + 1)$ - Jan 13th 2008, 10:12 PMjanvdl
- Jan 13th 2008, 10:15 PMMathMack
I got the part where you take the negative out, but after that i'm lost, I know it's (4a something)(a something) but lost to go after that, I know you must find factors but not sure

- Jan 13th 2008, 10:24 PMjanvdl
- Jan 13th 2008, 10:25 PMMathMack
and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!

2.) n^3-n^2+n-1

(n-1) (n^2 +1)

3.) 4a(a+2) - 8 (a+2)

4(a+2)(a-2)

4.) -6x^3 +10x^2 - 4x^4

-2x^2(2x+5)(x-1)

5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.

2x-5 - Jan 13th 2008, 10:27 PMjanvdl
- Jan 13th 2008, 10:30 PMMathMack
yes and on the first one we worked on they have the answer -(a+4) (4a-3)

Still lost on that one

THANKS JANVDLL! - Jan 13th 2008, 10:39 PMjanvdl
- Jan 13th 2008, 10:41 PMMathMack
that's what i'm not sure on either, the part that we should find two numbers for is positive 12

- Jan 13th 2008, 10:45 PMjanvdl
- Jan 13th 2008, 10:54 PMMathMack
okay, I finally got that one all figured out completly, just took awhile, The next one I can't figure out is the n^3-n^2 +n -1