what is an easy way to factor?

Question

such as:
12y+y^2+32
x^2+3x-40
20a-23a+6
i have my math final tomorrow and i dont know how to factor big problems such as this!
Also how do i know where to put the negatives and the positives in the final anwser?
please help! thanks!

Answer 1

I always begin by writing the equation with the squared term first:

12y+yexp2+32 becomes yexp2 + 12y + 32

In this case, there is no coefficient for the squared value, so I write the following: (y )(y )

I know the signs will be +/+ since both the middle and last terms are +: (y + )(y + )

Now list the factor pairs for the final term: 1*32, 2*16, 4*8

Find the pair, that when added together, equals the middle term coefficient: 1+32=33, 2+16=18, 4+8=12

Now put these in to the parens: (y+4)(y+8)

If you are given a problem with a coefficient on the squared term, do the same by listing all the factor pairs. Then determine your signs. If the last term is negative, then (y+ )(y- ); If last term is positive but middle term is negative, then (y- )(y- ).

Hope this helps!

Answer 2

I gave this answer for someone else's factoring question but it explains how to factor when the leading coefficient is not 1. I won't use your problems but this is an explanation of how to factor the first and third.

The expression to be factored here is 3x^2 + 16x +5. If you use this method and practice a lot, it become much, much easier to factor.


Michael Sullivan in his text book 'Algebra and Trigonometry' has a good method for factoring expressions where the coefficient of the first term is more than 1.
Let Ax^2 +Bx +C be the form so that A,B are coefficients and C is the constant.
Multiply A times C and get the product. Here it is 15 (call it AC).
Make a list of all the pairs of numbers which when multiplied together give you AC.
Here that list would be:
15*1
-15*-1
3*5
-3*-5
You have to include pairs of negative numbers when AC is positive because negative times negative produces positive.
Then look at the pairs. Which pair of numbers that has the product AC when ADDED together gives you B?
B is 16.
The only pair here that adds up to 16 is 15 and 1.
Rewrite the original expression:
3x^2 +15x +1x (this is your pair! and it is going to make the middle x term so we're talking here about quantities of xs.) +5.
The expression now reads 3x^2 +15x +1x +5
Now you factor by grouping: that is you try to make two groups of factors by finding common factors, first using the first term and another term and then pulling a common factor out of the group that's left.
You are aiming to end up with a factor that repeats.
In this case, the grouping gives you:
3x(x+5)+1(x+5)
You were looking for a factor that repeats. It is (x+5). The factors that you pulled out (3x and 1) make up a new factor.
Your result is (3x+1) (x+5)

Answer 3

the way i learned that was the box method.
for example:
y^2 + 12y + 32
Step 1: draw a 2 x 2 grid
Step 2: put y^2 in the top left box
Step 3: put +32 in the bottom right box
Step 4: find two numbers whose product is 32 ( i got this from the product of the 1 in front of the y and the 32 ) and their sum is 12 (the middle number)......which is 4 and 8.
Step 5: Put 4x and 8x in your other two boxes
Step 6: Pull out whatever is in common in both rows and both columns. So youll have x +4 on one side and x +8 on the other side.

I realize this is confusing with just words so I put a website that explains it better here: http://www.purplemath.com/modules/factquad2.htm