ok the question is how do i simplify the rational equation, If I can
15x^{2 }+ 24x + 9/ 3x + 3
ok the question is how do i simplify the rational equation, If I can
15x^{2 }+ 24x + 9/ 3x + 3
You take the same approach as with simplifying other rational expressions like... fractions!
Simplifying 15/6 is a no-brainer for most algebra students.
One method is to break up both the numerator and denominator into factors.
--> (3*5)/(2*3)
Obviously, you would cancel out the common factor, 3, giving you 5/2 as the simplest form.
Same principle with ratios of polynomials! You want to create factors and look to cancel out common factors.
15x^{2} + 24x + 9 can be factored into (3x+3)(5x+3)
**if you need explanation on how to factor this polynomial, see below.
So your factored rational expression now looks like (3x+3)(5x+3)/(3x+3)(1).
Notice that the numerator and denominator both have 3x+3 as a factor, so that factor will cancel out, leaving you with (5x+3)/1 or just 5x+3.
Answer: 5x+3
**help on factoring the trinomial in the numerator begins here:
If you have a trinomial that looks like ax^{2}+bx+c, it's called a quadratic trinomial.
I like to call the first and last term, ax^{2} and c, the "anchors."
To begin, start by thinking of factors for your anchors.
15x^{2} can be factored into 3x and 5x or
x and 15x.
9 can be factored into 3 and 3 or
1 and 9.
So now, you have a few possibilities. A few (not all) of which are:
(3x ± 3)(5x ±
3) or (3x ±
9)(5x ± 1) or (x ±
3)(15x ±
3) or (x ± 1)(15x ±
9)
Now all the above possibilities are based on your "anchors." Here's where your middle term comes in, your bx, or in this case, 24x. When you "FOIL" the above factors, you want the two middle terms to add up to 24x.
As an arbitrary example, say we used (3x+9)(5x+1). If we FOIL, we get 15x^{2}+3x+45x+9 = 15x^{2}+48x+9
Notice, our anchors are "good," but our middle term is not 24x. It's 48x, and that's way off! Now, you could try (3x+9)(5x-1), but that would give you 15x2-3x+45x-9. Now you have the wrong middle term AND the wrong "anchor." (your anchor was a positive 9, not a negative 9, remember?) In order to guarantee a positive 9, you will see that your factors will have to be (a+b)(c+d) or (a-b)(c-d) because (a+b)(c-d) will not give you positive anchors.
It does require some trial-and-error, but with perseverance, you'll eventually find that the correct factors are (3x+3)(5x+3).
Note: This method cuts down on a lot of trial and error, which makes factoring quicker/easier, but it is much easier to demonstrate in person. If you can learn this method, it'll make factoring trinomials like ax^{2} + bx + c much easier (when a > 1).
First, notice that all the terms in the numerator are divisible by 3, so you can pull out a factor of 3, and that you can pull a factor of 3 from the denominator, so we can rewrite the problem as:
3(5x^{2} + 8x +3)/3(x+1) and cancel off the 3's, to get:
(5x^{2} + 8x + 3)/(x + 1)
Now factorize the numerator (5x^{2} + 8x + 3). When you have ax^{2} + bx + c, the quickest way to factor it is to first multiply a by c, and then find two factors that will multiply to equal a*c (here, a*c = 5*3 = 15), add up to b (here, b = 8).
So, find factors that multiply to make 15, and add up to 8: 5*3 = 15 and 5+3 = 8. Then write the numerator like so, and you can pull a factor of 5 from the first set of parentheses:
(5x+5)(5x+3) = 5(x+1)(5x+3) but remember we multiplied a by c earlier, so we've actually multiplied the original numerator by a (a = 5), so we need to divide it by a, which we can do to get:
(x+1)(5x+3) and we've factored the numerator. Now refer to the green text above, which we can rewrite as:
(5x^{2} + 8x + 3)/(x + 1) = (x+1)(5x+3)/(x+1)
And now we can cancel off the (x+1) in the numerator and denominator to get: 5x+3