1. ## Factoring quadratic polynomial fractions

I have an additional question if someone would please help. If so, please explain how this. I simply do not understand how the numerator changes in this case.

$
\frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}
$

Expression SOLVED
$\frac{3}{2x^2+4x-48}+\frac{7}{6x-24}+\frac{1}{4x^2+20x-24}$

=

$\frac{3}{2(x-4)(x+6)}+\frac{7}{6(x-4)}-\frac{1}{4(x+6)(x-1)}$

The author then lists the correct LCD as $12(x-4)(x+6)(x-1)$. I agree with everything here except the coefficient of 12 should it not be

LCD = $2 * 6 * 4 (x-4)(x+6)(x-1)$ , where here I took the coeffecient factored out from each denominators expression and multiplied them by each other giving us a LCD of $48(x-4)(x+6)(x-1)$

it's 6+4+2 not 6*2*4

3. My memory must be foggy, are coefficients for the LCD always added in this manner?

4. Originally Posted by allyourbass2212
In this particular question the author wants me to find the LCD for the fractions.

Expression
$\frac{3}{2x^2+4x-48}+\frac{7}{6x-24}+\frac{1}{4x^2+20x-24}$

=

$\frac{3}{2(x-4)(x+6)}+\frac{7}{6(x-4)}-\frac{1}{4(x+6)(x-1)}$

The author then lists the correct LCD as $12(x-4)(x+6)(x-1)$. I agree with everything here except the coefficient of 12 should it not be

LCD = $2 * 6 * 4 (x-4)(x+6)(x-1)$ , where here I took the coeffecient factored out from each denominators expression and multiplied them by each other giving us a LCD of $48(x-4)(x+6)(x-1)$
In general you are right, but with this example the number 6 can be split into 2*3, that means the factor 4 is already present at:

$\underbrace{2\cdot 2}_{yields\ 4}\cdot 3 = 12$

5. I think it's easiest to think of it as 12 is the lowest number that 2, 6 and 4 are all factors of.

So you would multiply the top and bottom of the first fraction by $6(x-1)$ the second by $2(x+6)(x-1)$ and the third one by $3(x-4)$

6. Ah yes thank you very much earboth, small step I missed.

2=2*1
6=3*2
4= $2^2$

= $2^2*3$=12

7. I have an additional question if someone would please help. If so, please explain this $\frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}$, I simply do not understand how the numerator changes in this case.

8. Originally Posted by allyourbass2212
I have an additional question if someone would please help. If so, please explain how this $\frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}$, I simply do not understand how the numerator changes in this case.
do the two distributions in the numerator and combine like terms ...

$1(x-1) + 1(x+2) = x - 1 + x + 2 = 2x + 1
$

9. $1(x-1)+1(x+2)=x-1+x+2=2x+1$

*edit - Skeeter beat me to it