Hi Marie;
[4/(x^{2}6x+5)][(3/(x^{2}10x+25)]
Let's factor both denominators...
x^{2}6x+5
For the FOIL...
FIRST must be (x)(x)=x^{2}
OUTER and INNER must addup to 6x
LAST must be (5)(1) or (1)(5) and both numbers must be negative to render the sum of 6 and the product of +5.
(x5)(x1)
x^{2}10x+25
For the FOIL...
FIRST must be (x)(x)=x^{2}
OUTER and INNER must add up to 10x
LAST must be (5)(5) or (25)(1) or (1)(25) and both numbers must be negative to render the sum of 10 and product of 25.
(x5)(x5)
{4/[(x5)(x1)]}  {3/[(x5)(x5)]}
Let's take the first bracketed equation and multiply it by (x5)/(x5).
Let's take the second bracketed equation and multiply it by (x1)/(x1)
{[(4)(x5)]/[(x5)^{2}(x1)]}  {[(3)(x1)]/[(x5)^{2}(x1)]}
Now, the denominators are the same. Let's subtract...
[(4)(x5)][(3)(x1)]/[(x5)^{2}(x1)]}
(4x203x+3)/[(x5)^{2}(x1)]
(x17)/[(x5)^{2}(x1)]
Feb 15

Vivian L.