Easy Way To Factor Trinominals?

Question

I am currently in beginning algebra and we have reached the point of factoring trinominals. However, it seems to take FOREVER having to guess at which numbers work and which doesn't. Here is an example of a couple of my problems...

1. 16a^2 - 24a + 9

2. 36r^2 - 5r - 24

Any help on figuring out a more time-efficient way would be much appreciated. Thanks!

Answer 1

16a² - 24a + 9

The middle term is - 24a

find the sum of the middle term

Multiply the first term16 times the last term 9 equals 144 and factor

factors of 144 =

1 x 144
2 x 72
3 x 48
4 x 36
6 x 24
8 x 18
9 x 16
12 x 12. .<= . .use these factors

- 12 and - 12 satisfy the sum of the middle term

insert - 12a and - 12a into the equation

16a² - 24a + 9 = 0

16a² - 12a - 12a + 9 = 0. . .factor by grouping

4a(4a - 3) - 3(4a - 3) = 0

(4a - 3)(4a - 4) = 0

- - - - - - - - -s-

Answer 2

The AC method described is the best. If you practice this with a few trinomials with leading coefficients greater than 1, you will get so good at it, it only takes a minute or two to do one.
Basically, the method is to multiply the leading coefficient, in this case 16 (in the 16a^2-24a+9 example) by the constant which in this case is +9. Those are your A and C values. Their product is 144 so you look for a set of factors that multiplied together give you the product 144 but added together give you the B value which is the middle term (in this case, -24a).
144 is easy because you know 12 times 12 is 144 (a gross, 12 dozens, so it is easy to recognize this). It's just as easy to see that the two negatives, -12 and -12 multiplied together give you the positive 144 AND most importantly, added together, they add up to -24 (-12 plus -12 is -24).
The tricky part, if there is any, is looking at what happens when you write this all out:
You keep your first term and your last term in place and you substitute the -12a (you need to add two quantities of -12a together to get that middle term -24a) twice in the middle. Instead of writing -24a, you are substituting the equivalent, which you have chosen, '-12a -12a'.
So: 16^2 -12a -12a +9.
That was not the tricky part. The tricky part is looking at this and seeing how you can factor this by grouping so that you have two factors in parentheses that should be identical.
16a^2 - 12a is the first part of the equation that you wrote out and -12a + 9 is the second part of the equation. Look at each and see what you can do. 16^2a -12a can be factored with a 4a removed.
4a (4a-3)
Check by mulitplying out to get the first part back!
The second part is -12a + 9 and you look to see if you can get the 4a-3 in parentheses again if you factor. You can.
-3 (4a-3) Multiply it out to check.
Notice there is no 'a' outside the parentheses in this second part because when factoring here, you don't have an 'a squared' to deal with. That was already taken care of in the first part.
Now you have:
4a(4a-3)-3(4a-3)
Because the '4a-3' appears inside the parentheses both times, it has been distributed through and only is used once as a factor.
**At this point, if you're unclear on distributing, you'll get more confused. Review what distribution accomplishes. The distributive property, which works with multiplication and addition/subtraction, allows you to take a problem like [(6-2) times (4)] and rewrite it as [(6 times 4) - (2 times 4)] and the result is the same. You have to know this property, though; it's a basic.
The 4a and the -3 connect to make up the new factor 4a-3.
The two factors are (4a-3) and (4a-3). This is just (4a-3)^2.
It really is not that hard, once you can look at the expression and see how it can be grouped.

Answer 3

I've recently learned a new method called the "ac method". It involves multiplying a with c, and then finding two factors of ac which add up to the middle term, b. You then split the middle term in two, and use the technique of "grouping".

A reminder that a quadratic expression is in the form
ax^2 + bx + c

Let's employ that method for your two questions.

1. 16a^2 - 24a + 9

First, calculate ac. 16 x 9 = 144.

Next, find two factors of 144 that add up to -24 (which is b, the middle term's coefficient). They're obviously going to both be negative. The answer is -12 and -12.

Now, split up -24a into -12a and -12a. Look what happens.

16a^2 - 12a - 12a + 9

Factor the first two terms and the last two terms.

4a(4a - 3) - 3(4a - 3)

Now, group this expression by factoring (4a - 3) out of that.

(4a - 3)(4a - 3).

2. 36r^2 - 5r - 24

Step one: Calculate ac. 36 x (-24) = -864

Step 2: Find two factors of -864 which add up to -5.
With some effort, we can determine that -32 and +27 are the two factors which add up to -5.

36r^2 - 32r + 27r - 24

Factor the first two and last two terms.

4r(9r - 8) + 3(9r - 8)

Group by factoring (9r - 8) out.

(9r - 8) (4r + 3)

The convenient thing about this method is, regardless of how you split up your middle term, it will always group. For instance, had we split up -5 as 27r - 32 instead, we would get

36r^2 + 27r - 32r - 24
9r(4r + 3) - 8(4r + 3)

(9r - 8)(4r + 3)

Last note: Only worry about this when a is not equal to 1. When a = 1, i.e. you have something like x^2 - 8x + 12, factor by guessing.

Answer 4

1) (4a - 3) (4a - 3)

2) (9r - 8) (4r + 3)

I just do them through trial and error, there are only so many combinations to go through.

i.e for the second one, to get 36r^2, you can only use-:
36 and 1
18 and 2
12 and 3
9 and 4
6 and 6
You can write off the 36 and 1 straight away. I start with the lower values and go upwards.

I'm sure someone will have a technique to do them.

Answer 5

You can use a number a ways to find out the value(s) for a.

1)understand: Ax+Bx+C
you can use the Quadradic Formula

http://en.wikipedia.org/wiki/Quadratic_equation

-b+-square root of (b^2 -4ac)/2a

so with the first question using the quadradic formula
1. 16a^2 - 24a + 9
a)-(-24)+-(Sq rt. (-24^2-4*16*9))/2*16
b)24+-(Sq rt. (576-576)/32)
c)24+- (Sq rt. (0))/32
d)24/32 or 3/4

Answer 6

1.(4A-3)(4A-3)
=(4A-3)^2

2.(6R +8)(6R-3)