the problem is:

4x^2 + 12x + 9

the problem is:

4x^2 + 12x + 9

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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!

4x^{2} + 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 = 2.2.3.3 = 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 :)

4x^{2} + 12x + 9 = 4x^{2} + 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}

Molly,

Factoring by grouping (Matt) and factoring by FOIL (Arthur) are the two best ways for factoring trinomials regardless of whether the coefficient of the x^{2 }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.

Sincerely,

Krista K.

Molly,

There is a general procedure for factoring quadratic trinomials (ax^{2}+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 ax ^{2}+bx+c**

**GENERAL TECHNIQUE:**

- Step 1: Determine the product ac.
- Step 2: Search for factors of ac that add up to b. If they exist, call them m and n.
- Step 3: Rewrite the middle term of the quadratic as (m+n)x=mx+nx instead of bx.
- Step 4: Use
**factoring by grouping**to factor the new expression ax^{2}+mx+nx+c.

**EXAMPLE USING THE QUADRATIC YOU GAVE, 4x ^{2}+12x+9:**

- Step 1: Here we have a=4 and c=9, so ac=36.
- Step 2: We need to find factors of 36 that add up to b=12. Clearly 6 and 6 work.
- Step 3: Rewrite the quadratic as 4x
^{2}+(6+6)x+9=4x^{2}+6x+6x+9. - Step 4: Use
**factoring by grouping**to factor 4x^{2}+6x+6x+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, 4x^{2}+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:

4x^{2}+6x+6x+9=2x**(2x+3)**+3**(2x+3)**

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

2x(2x+3)+3(2x+3)

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, 2x ^{2}+5x+2:**

- Step 1: Here a=2 and c=2, so the product ac=4.
- Step 2: We need to find factors of ac=4 that add up to b=5. But clearly 1 and 4 work.
- Step 3: Rewrite the quadratic as 2x
^{2}+(1+4)x+2=2x^{2}+x+4x+2. - Step 4: Factor this new quadratic using
**factoring by grouping**.

Okay, now we have to carry out **factoring by grouping** on the quadratic

2x^{2}+x+4x+2

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

x**(2x+1)**+2**(2x+1)**

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

(x+2)(2x+1)

Hope my explanation makes the procedure clearer. Good luck in your math class!

Matt L

B.S. mathematics, MIT

Mollie, I use this method with my Algebra students all the time. It seems a bit complicated at first, but once you understand this technique, it is actually a great option to use when factoring trinomials with a coefficient is greater than one for the x squared term.

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)

And there are similar but more complicated formulas for roots of cubics and quartics, but no possible formula for degrees n>4: http://en.wikipedia.org/wiki/Abel_ruffini_theorem.

Molly,

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!

Art

4x^2 + 12x + 9 For this problem I would use the “ac method” that is: assume ax^2 + bx + c factors with integers. And multiply a by c then factor the result until the sum of the factors is equal to b. for our example a = 4, b= 12 c = 9. ac = a*c = 4*9 = 36 We start at the beginning with 1. 1*36 = 36, 1+ 36 = 37 oops! not 12 2* 18 = 36, 2 + 18 = 20 nope! not 12 3* 12 = 36, 3 + 12 = 15 nope! not 12 4* 9 = 36 , 4 + 9 = 13 man....??? still not 12 5* 7.2 = 36, 5 + 7.2 = 12.2 not 12 6*6 = 36, 6 + 6 = 12! AAALLL..RIGHTY Then!! so now we split b up into the factors we just found. Like so: in the original we have 12x so we split it up into (6 + 6 )x i.e. 6x + 6x. Then we write the original with the new b like this: 4x^2 + 6x + 6x + 9 Now re associate (put parenthesis where they'll do the most good!) (4x^2 + 6x) + (6x + 9) and factor out the least common multiple from each term 2x(2x+ 3) + 3(2x +3) Now factor out the common binomial term. [ 2x*(2x + 3) + 3*(2x + 3) ] = [2x*1 + 3*1 ] (2x + 3) = [2x + 3] (2x+3) so the factors are (2x+3)(2x+3) = (2x+3)^2.

Hi Eugene,

When you write near the end of your answer that we should "factor out the least common multiple from each term," what you mean is the
*greatest common factor (GCF)*, **NOT** the *least common multiple (LCM).
*The LCM of 4x^{2} and 6x is 12x^{2}. But the GCF of 4x^{2} and 6x is 2x.

It's easy to mix up these two ideas -- be careful!

Matt L

B.S. mathematics, MIT

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## Comments

Molly, Rekha is correct in stating that it helps to recognize factoring patterns in these types of problems. You do not want to have to re-invent the wheel every time you are faced with a factoring problem, especially later on when you are taking your ACT or SAT tests.

The goal is to quickly recognize if the problem can be solved easily through simple memorization; i.e. ask yourself, have I seen this problem before, or one very similar to it such that I can solve it in my head without much thought? For example, if the equation was instead 9x^2 + 30x + 25, you should be able instantly recognize the solution as (3x + 5)^2 using the a^2 + 2ab + b^2 = (a+b)^2 formula. If you can't solve it immediately, move on to the FOIL method as it takes less time to write everything down, then use factoring by groups, and finally the quadratic equation as a last resort to solve difficult problems.