Letf(x)=2x2+5x+1. If we writef(x) as f(x)=ax+1x-2+bx-2x-1+cx-1x+1 for real numbersa,b,c, then
There are infinite numbers of choices for a, b and c
Only one choice for a but infinite numbers of choices for b and c
Exactly one choice for each of a, b and c
More than one but finite number of choices for a, b and c
Explanation for the correct option:
f(x)=2x2+5x+1 and f(x)=ax+1x-2+bx-2x-1+cx-1x+1
Equating and comparing the coefficients
2x2+5x+1=ax+1x-2+bx-2x-1+cx-1x+1⇒2x2+5x+1=ax2-x-2+bx2-3x+2+cx2-1⇒2x2+5x+1=x2a+b+c+x-a-3b-c-2a+2b-c
This gives
a+b+c=2a+3b=-5-2a+2b-c=1
Solving the equations, we get
a=-4,b=-13,c=93
Hence, option (C) is the correct answer
Use the factor theorem to determine whether g(x) is a factor of f(x)
f(x)=22x2+5x+2;g(x)=x+2