# factoring trinomials

• Dec 16th 2007, 08:05 PM
n_duncan2010
factoring trinomials
Factor each trinomial , if not possible , write irreducible

Dont understand any of it, can anyone teach me how to do this???

1. 3x ^2 - 5x + 2

2. 5x ^2 + 17x + 6
• Dec 16th 2007, 10:29 PM
calculus_jy
hope it helps! very long
okay
if trinomial
it must be in the form (lx+a)(mx+b)
(x-a)(x-b) =lmx^2+(lb+am)x +ab
to obtain the left hand side from the right
we can do by (i) inspection or (ii) completing the sq
(i) we see that lm is the coefficient of x^2, (lb+am) of x and ab of x^0
therefore we find all the prime factors of lmab, then we can find 2 factors d, e such that de=lmab and d+e =lb+am
in number 1)3x ^2 - 5x + 2
lmab=6=2*3 lb+am=-5 therefore one d or e must be negative, but de =6 therefore both d and e must be negative
d e d+e
-6 -1 -7
-3 -2 -5 this is the one we want

so we do this
3x ^2 - 5x + 2 =3x ^2 - 3x -2x + 2
=(3x-2)(x-1)

we can do the step of finding d and e in our head with practise

(ii)3x ^2 - 5x + 2 = 3(x ^2 - 5/3x + 2/3)
we use the fact that (x+a)^2= x^2 +2ax +a^2
and here 2a= (5/3) a =5/6
3x ^2 - 5x + 2 = 3(x ^2 - 5/3x + 2/3)
= 3[(x-5/6)^2 -25/36+2/3]
=3[((x-5/6)^2 -(1/6)^2]
we use a^2 - b^2 = (a+b)(a-b)
3[((x-5/6)^2 -(1/6)^2]=3(x-1)(x-2/3)
=(3x-2)(x-1) same result
now u can try the 2nd one, usu we use the 1st method, but the second one always works but is very troublesome!
• Dec 17th 2007, 11:22 AM
n_duncan2010
thank youuu!!
:DThank You sooo!!!! much