When #p# is #-4<p<=3#, #(x+2)# is a factor of the function #f(x) =x^2-(p+1)x +p+4#. Moreover, #g(x) = xf(2x) - f(x - 1/3) +2/3x#. Determine the factors of #g(x)# without finding #f(2x)# and #f(x-1/3)#?

1 Answer
P dilip_k
Jul 13, 2016

#color(red)("Factors:- " x,(x+1),(12x-1)#

Explanation:

#color(blue)((x+2)" being a factor of "f(x)," the value of "f(-2) " should be zero.")#

So
#f(-2)=(-2)^2-(p+1)(-2)+p+4=0#

#=>4+2p+2+4=0#

#=>p=-10/3#

So #f(x)=x^2-(-10/3+1)-10/3+4#

#=>f(x)=x^2+7/3x+2/3#

#=>f(x)=1/3(3x^2+7x+2)#

#=>f(x)=1/3(3x^2+6x+x+2)#

#=>f(x)=1/3(3x(x+2)+1(x+2)#

#=>f(x)=1/3(3x+1)(x+2)#

#=>f(x)=(x+1/3)(x+2)#

This factorisation of f(x) reveals that (3x+1) and (x+2) are two fators of f(x)

So #color(red)(f(-1/3)=0" and "f(-2)=0#

Now we are to determine the factors of g(x) by trial,
where
#g(x)=xf(2x)-f(x-1/3)+2/3x#

Trial-1

#color(green)("Putting "x =-1->g(x)" we have")#

#g(-1)=-1*f(-2)-f(-1-1/3)-2/3#

#=>g(-1)=-1*0-f(-4/3)-2/3#

#= g(-1)=-f(-4/3)-2/3=#

Now
#=>f(x)=(x+1/3)(x+2)#

#=>f(-4/3)=(-4/3+1/3)(-4/3+2)#

#=>f(-4/3)=(-1)(2/3)=-2/3#

#:.g(-1)=-f(-4/3)-2/3#

#=-(-2/3)-2/3=0#

#color(red)("This means " (x+1)" is a factor of " g(x)#

Trial-2

#color(green)("Putting "x =0->g(x)" we have")#

#g(0)=0f(0)-f(0-1/3)+2/3xxo#

#=>g(x)=-f(-1/3)=0#

#color(red)("This means " (x)" is a factor of " g(x)#

Trial-3

On inspection it is found that g(x) contains a #x^3# as highest power of x. So another linear factor should exist.

#color(green)("Putting "x =1/12->g(x)" we have")#

#g(1/12)=1/12f(1/6)-f(1/12-1/3)+2/3xx1/12#

#=>g(1/12)=1/12f(1/6)-f(-1/4)+1/18#

Now

#=>f(1/6)=(1/6+1/3)(1/6+2)#

#=3/6xx13/6=13/12#

And

#=>f(-1/4)=(-1/4+1/3)(-1/4+2)#

#=1/12xx7/4=7/48#

#:.g(1/12)=1/12xx13/12-7/48+1/18#

#=(13-21+8)/144=0/144=0#

#color(red)("This means " (12x-1)" is a factor of " g(x)#

Related questions
Impact of this question
1152 views around the world
You can reuse this answer
Creative Commons License