Factoring?

It usually takes constant practice to be able to solve them without splitting. Nevertheless, I'll do my best and help you.

a. Let's first start with a quadratic with a =1

example: x^2 + x - 6, where a=1, b=1, c= -6

First create two brackets : ( ) ( )

then place x in the first positions of both brackets: (x )(x )

then find the two numbers that multiplies to get -6. i.e. 2 and 3

since b=1 (ie,its positive), the bigger one should be positive

therefore, we get

(x + 3)(x - 2)

try with x^2 + 5x - 6

if you get this right, then you are ready for the next challenge, a quadratic with a > 1

For example: 3x^2+17x+10, a=3, b=17, c=10

We still create our two brackets, but place 3x in the first and x in the second:

(3x )(x )

Why? Because the first term is 3x^2

Then find the two numbers that multiply to get 30 (remember to multiply a and c) and add up to get 17. That will be 15 and 2

Now, instead of fixing in the 15, remember that, when you expand, you'll have to come back to the same question, and since we already have 3x in one bracket, then we'll need just a 5

Therefore, we get

(3x + 2) (x + 5)

As I said, this comes by continuous practice. Try working as many examples as you can, and I know you'll get there. All the best!

I carry out steps as follows :-
(3x-------)(x---------)
(3x------2)(x------5)
(3x + 2)(x + 5)

3x^2 + 17x + 10
= (3x + 2)(x + 5)

3x^2 + 17x + 10
= 3x^2 + 15x + 2x + 10
= (3x^2 + 15x) + (2x + 10)
= 3x(x + 5) + 2(x + 5)
= (x + 5)(3x + 2)

SPlitting the term really is the best way. Honest. Any other method will just involve you creating a number of different possible answers and then checking each to see what works. This is a method of guessing and checking that will only work quicker if you are especially quick at mental math or really really lucky.

I'm assuming you know the method used to split the term -- in your example, you'd ask what two numbers mutilpy to 30 and add to 17? 15 and 2 and then you split the 17 into 15 and 2.

you have to have the middle term.

think of it like this:

(3x^2 + )(x+ )

what multiples of 10 do you plug in to get the original equation?

what multiplies into 10? there's 2 and 5, and 10 and 1.. odds are it's the 2 and the 5 so then plug it in and factor it out.

in this case it's:

(3x^2 + 2 ) (x + 5)

Yes you can use the formula which as follows:
x= { -b +(b^2 - 4ac)^1/2 }/2a
x ={ -b - (b^2 - 4ac)^1/2 }/2a

This formula will give you two values of x.