the problem is:
4x^2 + 12x + 9
the problem is:
4x^2 + 12x + 9
Usually in such algebraic equations, you need to see what pattern comes to mind. There are a number of formulae you should know by heart. Like the following :
(a+b)^2 = a^2 + 2ab + b^2
(a+b)(a-b) = a^2 - b^2
Similarly (a+b) ^3 = a^3 + 3a^2b + 3ab^2 +b^3
and others etc etc
If you now look at the equation it is in the form
(2x)^2 + 2.2x.3 +(3)^2 which is nothing but (2x + 3)^2.
Here we used the first formula (a+b)^2 = a^2 + 2ab + b^2 where
a = 2x, so a^2 = 4x^2
b= 3, so b^2 = 9 and
2ab = 2.2x.3 = 12x
I have always found knowing the formulae by heart helps a lot during factoring to help us recognize the patterns :)
Hope this helps!
4x2 + 12x + 9
Find all possible factors for 36 such that their product is 36 and sum ends up 12
Let us first distill it down to the lowest factors
36 = 4.9 = 126.96.36.199 = 2.3 X 2.3 = 6 X 6 (6+6 = 12!)
so 6 X 6 are the factors we want (other possible factors are 12X3, 36X1, 18X 2 etc, but none of them add up to 12)
The reason why we wanted to find the factors that add up to 12 is ?.... so that we can split the term 12x.. to do our magic work :)
4x2 + 12x + 9 = 4x2 + 6x + 6x + 9
Factor 2x from the first two terms and 3 from the second two terms (here I am looking for the common factors for both the first two terms and the last two terms. More can be explained but not explaning here)
2x (2x + 3) + 3 (2x + 3) ..There you go you have (2x +3) that you can factor now. So let us factro that out
(2x+3) (2x+3) OR (2x+3)2
Factoring by grouping (Matt) and factoring by FOIL (Arthur) are the two best ways for factoring trinomials regardless of whether the coefficient of the x2 term is 1 or greater. There is also a geometric way of solving these problems using squares and rectangles, but it tends to be a bit tedious. If you would like me to show you, I can, but it would need to be done in person.
There is a general procedure for factoring quadratic trinomials (ax2+bx+c), even those whose leading coefficient isn't 1. The method involves a special kind of factoring called factoring by grouping. Eugene P gave the basic idea, but he could have explained it better.
GOAL: Factor the quadratic trinomial ax2+bx+c
EXAMPLE USING THE QUADRATIC YOU GAVE, 4x2+12x+9:
To finish the problem, you have to carry out Step 4 -- so you have to know how factoring by grouping works. The basic idea is to group the first two terms together (in the example above, 4x2+6x) and the last two terms together (in the example above, 6x+9), and then factor each of those separately. In the example above, we can factor out a 2x from the first two terms and a 3 from the last two terms. This gives us:
Now the whole point to this procedure is that the same factor, (2x+3), shows up in each group (that's why I've bolded them above). So you can, in turn, factor IT out of the sum
to get the final answer of (2x+3)(2x+3)
This problem is slightly special because your quadratic turns out to be a perfect square (the 2x+3 is a repeated factor), but this need not be true in general.
ANOTHER EXAMPLE, 2x2+5x+2:
Okay, now we have to carry out factoring by grouping on the quadratic
If we group the first two terms and the last two terms, we can take an x out of the first two and a 2 out of the last two. That gives us
Now the whole point is that the (2x+1) shows up for both groups (that's why I've bolded them above), and we can, in turn, factor IT out, in which case we get
Hope my explanation makes the procedure clearer. Good luck in your math class!
B.S. mathematics, MIT
You can also learn the quadradic formula. It will never let you down, and you have to learn it eventually. :) Check out the video on you-tube for a memory trick/song.
http://www.youtube.com/watch?v=CnJT1ojHT28 (jingle bells tune)
Use the foil method. You know that two possible combinations will produce 4x^2
4x * x or 2x * 2x
You know that the last term factors could be 9*1 or 3*3
The real question is what will produce the middle term? 4*3 = 12 as well as 12*1 and 6*2.
Notice that 4 ans 3 are common to the first and last term ( I say this becuase 3 is in 3^2 and in 6 which is 2*3) factors and to the middle tern but then if you attempt:
(4x+3)(x+3) the middel term would be 15 so that means that we are left with 2 and 2 for the first term factors such that
(2x+3)(2x+3) which would produce 4x^2 + 6x +6x +9 or 4x^2 +12x +9
You could also use the quadratic formula where:
[-12 +/-(12^2 -4(4)(9))^1/2]/2(4) = [-12+/-(144-144)^1/2]/8 = (-12+0)/8 or -(12-0)/8 = -12/8 = -3/2
Note that if x = -3/2 then 2x = -3 and 2x+3=0 and since there are two solutions then (2x+3)(2x+3) are the factors of the quadratic equation.
Hope this helps!