Expert: Scott A Wilson - 7/10/2010

Question
I really need help on my complete factoring
while u solve it can you also explain how to do it?
thanks

1. 6x²-11x+5
2. 64x²-16x-1
3. 25y^4-144z²
4. 2x²+10x-168
5. x^6-7x^4+12x²

Answer
Note what the factors are of the first number and the last.
The sum of the product is the same as the middle if both terms have the same sign
and the difference of the product is that middle term if one of them is negative.

1. For this one, the factors of the first constant are 1*6 or 2*3.
The factors of the last constant are 1*5.
They are of the same sign, and the middle terms has an 11.
This means 11 must be gotten as 1*1 + 6*5, 1*5 + 6*1, 2*1 + 3*5, are 2*5 + 3*1.
The totals of these four expressions are 1+30=31, 5+6=11, 2+15=17, are 10+3=13.
Wnce only the second one works, we must have a 5 and a 6 involved, which means it is
(6x+5)(x+1).  The first term in each expression gives us 6x² and that last term in each expression gives us 5, which is what we need.  To get the middle term, multiply the first in the first expression by the last in the last and then the last in the first by the first in the last.

I think of this term as being gotten by multiplying the two outside together and the two inside together, which gives us 6*1 + 5*1 = 11.

2. The last term is a -1, and the factors of the first term are 1*64, 2*32, 4*16, or 8*8.
Since the last term is a negative, the difference of two of these terms must be 16.
This is not so, but if the last term were a +1, it would be (8x-1)(8x-1).

Beyond what you know right now:
There is a quadratic equation which says for ax²+bx+c=0, the factors are (-b±√(b²-4ac))/(2a).
Using the numbers given as a=64, b=-16, and c=-1, this gives (-16±√(256+256))/128.
Since the √256 = 16, this is 16(-1±√2)/128 = (-1±√2)/8.
The terms are then 64(x+a)(x-b), where a = (-1+√2)/8 and b = (-1-√2)/8.

3. Since 25 = 5² and 144 = 12², this is the difference of two squares.
It is known that for such a difference, a² - b², the factors are (a-b)(a+b).

4. The factors of the first term are 1 and 2.
Since there's a minus, the product of 1 and 2 with factors of the last term needs to be 10.
The last of the last term are 2*84, and 84 is 2*42, so they are 2*2*42.  Now 42 is 2*21, so they are 2*2*2*21.  Lastly, 21 is 3*7, so they are 2*2*2*3*7.

This means the last term is 1*168, 2*84, 3*56, 4*42, 6*28, 7*24, 8*21, or 12*14.
Since the difference is 10, find a set such that two times one minus the other gives 10.
168 - 2*1 = 166, so that's too large.
84 - 2*2 = 80, so that's too large
56 - 2*3 = 50, and that's too large.
42 - 2*4 = 35, and that's getting close, but still too large.
28 - 2*6 = 16, and that's really close, but still too large.
24 - 2*7 = 10, and bingo! ... that's it!
This means to put the 1 and the 24 opposite each other and the 2 and 7 opposite each other.
This gives (x-7)(2x+24).  Checking the outside and inside, we get 24-14=10, and that's right.

5) First, factor out x² giving x²(x^4-7x²+12).
Next, replace x² with a, so I say a=x² and make the replacements.  That gives a(a²-7a+12).
Now that factors of the last term are 1*12, 2*6, or 3*4.
Adding them together gives 13, 8, and 7.
Since we are looking for the 7, and it is negative, this means that it is a(a-3)(a-4).
Since a=x², that can be put back in.