An important question from factorisation chapter for examination as it has been already asked in previous paper in 2012 in which we are to factorise completely the following polynomial 3x3+2x2−19x+6 using remainder theorem
ML Aggarwal Avichal Publication Factorisation chapter 6, question no 16(ii)
Let f(x)=3x2+2x2−19x+6
Using hit and trial method,
f(1)=3+2−19+6=0
f(−1)=−3+2+19+6=0
f(2)=24+8−38+6=0
∴(x−2) is a factor of f(x).
3x2+8x−3
Now, x−2)3x2+2x2−19x+6
3x2−6x
———————————-
8x2−19x
8x2−16x
————————————
−3x+6
−3x+6
————————————-
x
————————————–
To factorise 3x2+8x−3
=3x2+9x−x−3
=3x(x+3)−1(x+3)
=(3x−1)(x+3)
Hence 3x3+2x3−19x+6=(x−2)(3x−1)(x+3)