When you're factoring polynomials you're taking a sum of terms (for example 6 x ^ 2 - 2 3 x + 1 5) and turning them into a product of factors (x+?)(x+?). If we have ax^2+bx+c we can turn that into (x+m)(x+n) where m and n are extra constants. This is just like factoring a real number into prime factors. For instance the number 56 can be factored into smaller factors
Which closely represents (x+m)(x+n), say we let x=5, m=3,n=2 then
it comes to (5+3)(5+2)=(8)(7).
This product can be written as a sum also. Say we knew m and n, but not x. We couldnt mix the x's and the numbers, but we can FOIL it.
Now we have a polynomial.
In this case we're going to work backwards: we're going from a sum (6x^2 - 23x + 15) to a product (cx+m)(dx+n) (where c and d are equal to 1 sometimes) which will FOIL to cdx^2+cnx+dmx+mn (or sometimes x^2+nx+mx+mn). The result cdx^2+cnx+dmx+mn is equal to 6x^2 - 23x + 15 and you can see the terms line up.
6x^ 2 - 23 x + 15
cdx^2 +(cnx+dmx) + mn
The last term of 15 is equal to mn (ie mn=15), cdx^2=6x^2, and cnx+dmx=-23x. The trick is to find c,d,m, and n. The last term mn is equal to 15 so our factors could be 1,3,5,15 or -1,-3,-5,-15 (in other words: say m=3,n=5 mn=3*5=15. This is possible for any combination of factors that multiply to 15). To find c and d, notice how cdx^2=6x^2. If we get rid of the x^2 it comes to cd=6, so the factors of 6 are 1,2,3,6.
Also since the middle term is negative we know that cnx+dmx is negative. Since the last term is positive and the middle term is negative, m and n must both be negative. If I multiply two negative numbers, I get a positive number; If I add two negative numbers, I get a negative number. So lets try a few combinations of these factors: let c=1,d=6,m=-15,n=-1
We got 2 of the 3 terms of the polynomial, but instead of -91 we need -23 for the middle term. So lets try again, let c=1,d=6,m=-5,n=-3:
Again we got everything but the middle, so lets try another pair let c=1,d=6,m=-3,n=-5:
We finally got it (well I already knew the answer, if I didn't know it I would have to try more combinations). To find the right combo it takes a lot of practice and a bit of luck. In short, 6x^2-23x+15 factors to (x-3)(6x-5).
Now take this product and set it equal to zero.
This resembles ab=0 where a=(x-3) b=(6x-5). To solve for a, divide both sides by b (or 6x-5)
Now we can find x
There's one answer, do the same thing but find b
There's our other answer. These 2 answers are the roots (x-intercepts) to the polynomial. If we plug in x=3 or x=5/6 into (x-3)(6x-5) or 6x^2-23x+15 we will get zero. I hope at least one sentence of this makes sense (I'm sure you got most of it) so feel free to ask even more questions after this explanation. If you want a quick way, graph it and find the x-intercepts.
The solution represents the roots