# Factoring Pattern for ax^2 + bx + c

1. Feb 19, 2005

### IndigoSwing4

I'm currently having trouble solving some problems that are in the form ax^2 + bx + c. I can get the factors but its just the process of putting the factors into two binomals that I'm having trouble with. If someone could clearly explain how to do this, it would be greatly appreciated. Down below I have two example problems from my work that pretain to this.

1. 6h^2 + 2h - 1

2. 2d^2 - 7d + 6

2. Feb 20, 2005

### matt grime

IF it factors easily then a simple analysis like this will work:

Suppose (ah+b)(ch+d) = 6h^2 + 2h -1, then bd=-1, so supposing they are integers, we can take b=1, d=-1

ie (ah+1)(ch-1) = 6h^2 + 2h -1, Then ac=6, so again assuming integer answers, a=6, c=1, a=2, c=3, a=3, c=2, etc. Look at the cross terms now and see what happens.

Note this isn't guaranteed to work.

We can do other things though, such as completing the square or using the formula (which is actually just noting that completing the square is the same every time)

so 6h^2+2h-1 = 6(h^2 + h/3) -1 = 6(h +1/6)^2 - 1/6-1 = 6(h +1/6)^2 - 7/6

setting equal to 0 to find the roots

6(h +1/6)^2 - 7/6=0

(h+1/6)^2 = 7/36

h+1/6 = +/- sqrt(7)/6

So in this case we see that trying to spot the roots by looking at factors would've led you nowhere, really.

3. Feb 20, 2005

### mathwonk

the only factoring trick we understand well usually is the difference of two squares.

i.e. x^2 - a^2 = (x-a)(x+a). if you do not know this one, learn it immediately, you will never regret it. without it you are totally helpless.

now suppose we have any quadratic like x^2 + bx + c.

then we try to turn it into a differrence of two squares as follows:

x^2 + bx + (b/2)^2 -(b/2)^2 + c = [x+ (b/2)]^2 - [(b/2)^2 - c]

this works provided the last bit is a square: i.e. if [(b/2)^2 - c] is a square.

if so wee call its square root sqrt([(b/2)^2 - c])

and then we have that x^2 + bx + c = 0 if and only if

[x+ (b/2)]^2 - [(b/2)^2 - c] = 0,

if and only if ([x+ (b/2)] - sqrt([(b/2)^2 - c])( [x+ (b/2)] + sqrt([(b/2)^2 - c]) = 0.

so the two roots are x = - (b/2)] ± sqrt([(b/2)^2 - c])

= -b/2 ± sqrt(b^2 - 4c)/2. the usual formula (when a = 1).

forgive me if i am not answering any question you actually have.

4. Feb 20, 2005

### James R

IndigoSwing4,

Let's look at one of your examples.

1. 6h^2 + 2h - 1

Look at the first and last terms: 6 and -1. What are the factors of each?
For 6, we have: (1 times 6) or (2 times 3).
For -1, we have only 1 times -1.

Now look at the middle term, which in this case is +2. We need to multiply one of the factors of 6 by one of the factors of -1, then multiply the two other factors of 6 and -1 and finally add the two multiples together to get +2.

Suppose we choose the factors 6 and 1. That leaves the factors 1 and -1. Multiply the two pairs and add gives: $6 \times 1 + 1 \times -1 = 5$, which is not +2, so that doesn't work.

Instead, choose the factors 3 and 1. This time we get $3 \times 1 + 2 \times -1 = +2$, which works.

Therefore, the correct factorisation is:

$$6h^2 + 2h - 1 = (3h - 1)(2h + 1)$$

Notice, the 3 is multiplied by the +1, and the 2 is multiplied by the -1. The two then add to give +2.

Hope this helps!

5. Feb 21, 2005

### Muzza

The only problem is that (3h - 1)(2h + 1) is actually equal to 6h^2 + h - 1.

6. Feb 22, 2005

### James R

Doh!

Well, I hope the method helped, anyway.