Can someone explain this?
Factoring 6r2-23r+20
The first term is, 6r2 its coefficient is 6 .
The middle term is, -23r its coefficient is -23 .
The last term, "the constant", is +20
Step-1 : Multiply the coefficient of the first term by the constant 6 • 20 = 120
Step-2 : Find two factors of 120 whose sum equals the coefficient of the middle term, which is -23 .
| -120 | + | -1 | = | -121 | ||
| -60 | + | -2 | = | -62 | ||
| -40 | + | -3 | = | -43 | ||
| -30 | + | -4 | = | -34 | ||
| -24 | + | -5 | = | -29 | ||
| -20 | + | -6 | = | -26 | ||
| -15 | + | -8 | = | -23 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -15 and -8
6r2 - 15r - 8r - 20
Step-4 : Add up the first 2 terms, pulling out like factors :
3r • (2r-5)
Add up the last 2 terms, pulling out common factors :
4 • (2r-5)
Now add up the four terms of step 3 :
(3r-4) • (2r-5)
Which is the desired factorization
+2353 Hey Chris,
I think the person that asked this question actually wanted an explanation why this works 
.
Now it does somehow seem logical in my head, but I'm too tired to be able to put it on paper.
Will you have another look at it?
Otherwise I'll have a look at it tomorrow or the day after tomorrow... 
+129852 Suppose we have the quadratic....... ax^2 + bx + c
This technique will work as long as we can find another factorization of ac where the factors sum to "b"
To see why, let's suppose ac = n
Now, let us suppose that there are two other factors of n, say q and r, such that q + r = b....then...
(a must equal q*m) and (c must equal r/m) because (qm) * (r/m) = q*r = n
So, we have..........
ax^2 + bx + c =
(q*m)x^2 + (q + r) x + (r/m) =
(q*m)x^2 + qx + rx + (r/m) =
qmx(x + 1/m) + r (x + 1/m) =
(qmx + r) ( x + 1/m)
Which shows that this type of polynomial can be factored as long as ac - i.e., n - has factors q,r such that q + r = b
+118677 I use this method all the time! It is the BEST way to do it!
Factoring trinomials using the grouping method.
This is a really good clip.
https://www.youtube.com/watch?v=HvBiJ9W00Z4
If you have any questions just ask.
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(I'd like to point out to other answerers that this is just one of the addresses included in my
"Information pages worth Keeping and Developing" thread - It has been there for a while!
