While the $\displaystyle (x-2)$ term in the denominator might be off-putting, let's start with the numerator and see where things go. We can try and factor the numerator by finding two numbers that add up to $\displaystyle 1$, and multiply to get $\displaystyle -6$.  It just happens that $\displaystyle 3$ and $\displaystyle -2$ do that.  Now our equation looks like this:

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

We can now cancel the $\displaystyle (x-2)$ from the denominator for:

$\displaystyle f(x)=x+3$

And then solve:

$\displaystyle f(3)=(3)+2$

$\displaystyle f(3)=5$

$\displaystyle x^2-4x+4$

Check if the problem factors using the following property,

$\displaystyle x^2+ cx+d = (x+a)(x+b)$

where c will be the sum of a and b, and d will be the product of them. We can prove this by expanding out the right hand side:

$\displaystyle (x+a)(x+b)$

$\displaystyle =x^2+ax+bx+a*b$

 $\displaystyle =x^2+(a+b)x+(a*b)$

Therefore: $\displaystyle (a+b) = c$    and:  $\displaystyle a*b = d$

In our problem, -4 is the sum of a and b, and 4 is the product of a and b. What numbers for a and b can we use to satisfy our equation? Well, the numbers will have to be factors of 4, so we can start by writing these down: 1*4   and 2*2 Trying these out we can clearly see that the proper numbers are 2, and 2, since they are the only ones that add up to 4! Now that you have a and b, plug these back into the original relationship to find:

$\displaystyle (x-2)(x-2)$

Expand this out for yourself to see that the property holds true!

$\displaystyle x^2+10x+24$

Check if the problem factors using the following property,

$\displaystyle x^2+ cx+d = (x+a)(x+b)$

where c will be the sum of a and b, and d will be the product of them. We can prove this by expanding out the right hand side:

$\displaystyle (x+a)(x+b)$

$\displaystyle =x^2+ax+bx+a*b$

 $\displaystyle =x^2+(a+b)x+(a*b)$

Therefore: $\displaystyle (a+b) = c$    and:  $\displaystyle a*b = d$

In our problem, 10 is the sum of a and b, and 24 is the product of a and b. What numbers for a and b can we use to satisfy our equation? Well, the numbers will have to be factors of 24, so we can start by writing these down: 1*24  2*12   3*8    and 4*6. Trying a few of these out we can clearly see that the proper numbers are 4, and 6, since they are the only ones that add up to 10! Now that you have a and b, plug these back into the original relationship to find:

$\displaystyle (x+4)(x+6)$

Expand this out for yourself to see that the property holds true!

Here we would need two numbers that multiply to get to $\displaystyle 9$, but would cancel each other when added.  That means that the number is going to be the same, but one will be positive and one will be negative.  Looking through the factors of $\displaystyle 9$, $\displaystyle (1, 3, 9)$, we see that $\displaystyle 3$ multiplied by $\displaystyle 3$, gives us $\displaystyle 9$, which is what we want.  We then make one $\displaystyle 3$ positive, and the other negative:

$\displaystyle f(x)=(x+3)(x-3)$

Here we start to factor by first grouping the terms together:

$\displaystyle f(x)=(x^{3}+3x^{2})+(3x+9)$

From there we can see that we can factor an $\displaystyle x^2$ out of the first group of terms, and a $\displaystyle 3$ from the second group of terms:

$\displaystyle f(x)=x^{2}(x+3)+3(x+3)$

Now, because both groups share a common $\displaystyle (x+3)$, we can factor that out of both of them:

$\displaystyle f(x)=(x+3)(x^{2}+3)$

Just to make sure we're finished, we can check to see if any of the remaining groups can be factored.  None can, so we're done.

To factor, let's begin by grouping the first two terms and the last two terms:

$\displaystyle f(x)= (x^3 + 2x^2) +(6x +12)$

We can then factor out a common $\displaystyle x^2$ from the first grouping, and a $\displaystyle 6$ from the second grouping:

$\displaystyle f(x)= x^2(x+2)+6(x+2)$

As we can see, there's a common $\displaystyle (x+2)$ from each group, and we can factor that out as well:

$\displaystyle f(x)=(x+2)(x^2 + 6)$

From here, we can double check that none of the terms can be factored further.  They can't be, so we're done.

To begin, let's group the first two terms:

$\displaystyle f(x)=(x^{3}+x^2) -x -1$

We can easily see that we can factor out a $\displaystyle x^{2}$ from the first group, giving us:

$\displaystyle f(x)=x^2(x+1) -x -1$

What's tougher to see is that we can also factor out a $\displaystyle -1$ from the second grouping, giving us:

$\displaystyle f(x)=x^2(x+1) +(-1)(x +1)$

Now we have a $\displaystyle (x+1)$ in both groups, which we can factor it out:

$\displaystyle f(x)=(x^2 - 1)(x+1)$

We're not quite done yet though.  We can still factor:

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

Giving us:

$\displaystyle f(x)=(x+1)(x+1)(x-1)$

We then make things look slightly nicer by collecting all the like terms for a final answer of:

$\displaystyle f(x)= (x+1)^2 (x-1)$

We can split this problem into two different parts.  The first part is to find the greatest common constant (the numbers our in front of each term), and the greatest common variable (the $\displaystyle x$).  The variable part is much easier; each term has an $\displaystyle x$ in it, and the limiting term is the last one (it only has 1 $\displaystyle x$), so we can factor it out.  That leaves us with:

$\displaystyle f(x)=x(24x^{3}+12x^2 +57)$

Looking at the numbers, we can see that each has $\displaystyle 3$ as a factor, so we can also factor that out:

$\displaystyle f(x)=3x(8x^{3}+4x^2 +19)$

At this point, there's nothing else that we can factor out, so the largest common factor (that we were able to remove from each term) was:

$\displaystyle 3x$

We can start by only worrying about the numerator, and grouping the first two and the last two terms together:

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

From here we can factor out a $\displaystyle x^2$ from the first term, and a $\displaystyle -4$ from the second term:

$\displaystyle x^2(x+1)-4(x+1)$

We can now factor out the $\displaystyle (x+1)$ from each term:

$\displaystyle (x^2 - 4)(x+1)$

While we could continue to factor the $\displaystyle (x^2 - 4)$ term, let's see what the problem looks like overall for now:

$\displaystyle f(x)= \frac{(x^2 - 4)(x+1)}{x^{2}-4}$

We can cancel the $\displaystyle (x^2 - 4)$ from the numerator and the denominator to save some time.  If we were to factor everything out and then cancel, we would still get the same answer:

$\displaystyle f(x)=(x+1)$

$\displaystyle x^2+7x+12$

Check if the problem factors using the following property,

$\displaystyle x^2+ cx+d = (x+a)(x+b)$

where c will be the sum of a and b, and d will be the product of them. We can prove this by expanding out the right hand side (Using FOIL!):

$\displaystyle (x+a)(x+b)$

$\displaystyle =x^2+ax+bx+a*b$

 $\displaystyle =x^2+(a+b)x+(a*b)$

Therefore: $\displaystyle (a+b) = c$    and:  $\displaystyle a*b = d$

In our problem, 7 is the sum of a and b, and 12 is the product of a and b. What numbers for a and b can we use to satisfy our equation? Well, the numbers will have to be factors of 12, so we can start by writing these down: 1,12  2,6, and 3,4. Trying a few of these out we can clearly see that the proper numbers are 3, and 4, since they add up to 7! Now that you have a and b, plug these back into the original relationship to find:

$\displaystyle (x+3)(x+4)$

Expand this out for yourself to see that the property holds true!