Question 217206
To reduce/simplify fractions we need to express the numerator and denominator in factored form. Then, if there are factors comon to the numerator and denominator, we can cancel these factors.<br>
1) {{{(4y)/(2xy + 4y)}}}
The numerator is already in factored form (i.e. it is a multiplication). In the denominator we can factor out the Greatest Common Factor (GCF) which is 2y:
{{{(4y)/(2y(x+2))}}}
To make the common factors more obvious, I'll factor the 4, too:
{{{(2*2y)/(2y(x+2))}}}
Now we have a common factor, 2y, which we can cancel:
{{{(2*cross(2y))/(cross(2y)(x+2))}}}
{{{2/(x+2)}}}
And we are finished. (Note: You cannot cancel the 2's! The 2 in the denominator is <b>not</b> a factor and <b>only</b> factors can be canceled.)<br>

2) {{{(2x^2 + 4x)/(2x^2 +6x)}}}
Again, we start by factoring. In the numerator the GCF is 2x and in the denominator the GCF is also 2x:
{{{((2x)(x + 2))/((2x)(x+3))}}}
The (2x)'s cancel leaving:
{{{(x + 2)/(x+3)}}}
(The x's do not cancel because they are not factors.)<br>

3. {{{(x^2 + 4x +4)/(x + 2)}}}
This time there is no GCF (other than 1) in either the numerator or denominator. But GCF is not the only kind of factoring. One of the other methods of factoring is factoring by patterns. One of the patterns is: {{{a^2 +2ab + b^2 = (a+b)(a+b) = (a+b)^2}}}. The numerator fits this pattern because {{{x^2 + 4x +4 = (x)^2 + 2(x)(2) + (2)^2}}}. (Look at this until it makes sense. See how the two expressions are equal. See also how the expression on the right also fits the pattern {{{a^2 +2ab + b^2}}} with "a" being "x" and "b" being "2".) So factoring the numerator with this pattern we get:
{{{((x+2)(x+2))/(1*(x + 2))}}}
Now one of the (x+2)'s on top cancel the x+2 on the bottom leaving simply:
{{{x+2}}}<br>

4. {{{(4x^2 - 12x -40)/(2x^2 - 16x + 30)}}}
Whenever you factor start with the GCF (if it is not 1). Then try any and all other methods until you can no longer factor anything. The GCF on top is 4 and the GCF on the bottom is 2. Factoring these GCF's out we get:
{{{(4(x^2 - 3x - 10))/(2(x^2 - 8x + 15))}}}
Now we try other factoring methods. The method that works on the numerator and denominator in this problem is trinomial factoring (with a=1). This is where you ask and answer the question, "What factors of the constant term (at the end) add up to the coefficient in the middle?" For the numerator this question would be: "What are the factors of -10 that add up to -3?" After some thought I hope you see that the answer is -5 and 2. So the {{{(x^2 - 3x - 10)}}} factors into {{{(x-5)(x+2)}}}.
The question for the denominator is "What are the factors of 15 that add up to -8?". This one is a little tricky. You have to realize that in order to add up to -8 at least one number must be negative. And in order for the numbers to multiply to a positive 15 both numbers must be positive or both numbers must be negative. Combining these we learn that both factors must be negative and once we realize this we can find that they must be -3 and -5. So the {{{x^2 - 8x + 15}}} factors into {{{(x-3)(x-5)}}}
Putting this all together we now have:
{{{(4(x-5)(x+2))/(2(x-3)(x-5))}}}
or
{{{(2*2(x-5)(x+2))/(2(x-3)(x-5))}}}
The (x-5)'s cancel and one of the 2's on top cancel with the 2 on the bottom leaving:
{{{(2(x+2))/(x-3)}}}
This will no reduce any further. But we can multiply out the numerator:
{{{(2x+4)/(x-3)}}}<br>
As you can see, reducing fractions like these requires that<ol><li>you know factoring must be done first.</li><li>you are good at factoring. There are many methods of factoring and it takes practice to learn them and get good at them.</li></ol>