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!