How do we simplify radical expressions?

           Radicals are by far the scariest looking thing ever. Seriously I mean look at this for example, √18x⁵y⁴z . How the heck do you solve that? Well I'm here to help you. Lets take this one step at a time. First step ignore the radical it makes your life easier. Even though 18 is not a perfect square, it can be broken down into smaller pieces where one of those pieces might be perfect square. From 18 we can say 9 and 2 multiply to give you 18 and 9 is a perfect square. From x⁵ we get x⁴ and x. We leave y⁴ alone because its already "perfect" . And z is all alone so it just stays z. 
       So now under the Radical we have √9•2•x⁴•x•y⁴•z. Now we begin to just pull out the perfect parts we take out nine and that turns into a 3 outside of the radical. We can't do anything with two so it stays inside. x⁴ comes out as x² and the single x stays inside. With y⁴ we bring it out and it turns into y² with no y's inside. Lastly x was all alone and it remains that way inside the radical. 

        Now we just take our perfect outside piece, 3x²y² and combine it with what stayed in our radical √2xz and our final answer is 3x²y²√2xz. Suddenly simplifying these radicals seems like a breeze! All we did was this:                   

           Not that hard huh?

How do you factor a trinomial where a>1 ?

                    Factoring by grouping was simple to understand but, what if now you get a trinomial that looks like: 6x2+19x+10? In this case factoring by grouping will not work but there is a way to factor this easier than you would expect.

            First you take your a and your c, in this case 6 and 10, and multiply them together giving you 60. Now you will find two numbers that multiply to give you your product of a and c and add to give you b. So were looking for two numbers that multiply to give you 60 and add to give you 19. If you guessed 15 and 4, you got it!

          Now you take these two numbers and replace them in your old equation. Your new equation will look like: 6x2+15x+4x+10. Now what do you do? Well now you can factor by grouping. You know the drill, Take the first two terms and the last two terms and pull out the greatest common factor. The GCF of the first two terms is 3x and the GCF of the last two terms is 2. So currently you have 3x(     )+2(     ).

                 You'll bring down what you have left which will be 2x+5 in both equations. Now that they have something in common you combine both 2x+5's and combine the two GCF's leaving you with a final answer of (3x+2)(2x+5). Bing bang bomb you factored that dude. 

How do you factor by grouping?

            You're taking a test and you're doing great all of a sudden you get to a question that says Factor: 2x3+4x2 +5x+10. In your head you think "crap" but have no fear! I'm hear to show you how it's done! 

       First split up the expression into two parts. The first two and the last two. Deal with both parts alone. Now working with the first part pull out the GCF. The GCF of 2x3+4x2 is 2x2 now you bring down whats left which is (x+2). When you put that all together what you get for the first half is        2x² (x+2).

         Now you do the same thing to the second half of the expression. Pulling out the GCF of 5x+10 gives you 5. Pulling out whats left is going to give you (x+2). Notice that after taking out the GCF both half's of the equation have the same thing left over. Now you combine everything for the second half givig you 5(x+2). 

          
        You have now factored both half's of the equation and its time to bring it all to together. right now your parts look like 2x²(x+2)+5(x+2). (x+2) is common on both sides so you can combine that. Now you combine both GCF's and your final answer looks like (2x²+5)(x+2). Now you can finish your test without losing your mind.