Can you find the values of 'a' and 'b' using the remainder theorem?

Question

The remainder is #5# when #P(x)=ax^3-4bx^2+x-4# is divided by #x-3# and the remainder is #2# when #P(x)# is divided by #x+1#. find the values of #a# and #b#.

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Answer 1 · 2018-03-27T12:21:11

#a = -19/12#

#b = -195/144#

The fact that you are asked to solve for #a and b# indicates that we will be working with two equations and solving them simultaneously,

Given #P(x) = ax^3-4bx^2+x-4#

Using the remainder theorem we can say that :

if #x = 3#, then the remainder is #5#

#P(3) = a(3)^3-4b(3)^2+(3)-4=5#

#27a-36b -1=5#

#color(red)(27a-36b = 6)#

if #x =-1#, then the remainder is #2#

#P(-1) = a(-1)^3-4b(-1)^2+(-1)-4=2#

#-a-4b-1-4=2#

#color(blue)(-a-4b = 7)#

Solve the equations simultaneously

#color(blue)(-a-4b = 7)" "rarr a = -4b-7#

#color(red)(27a-36b = 6)#

#color(red)(27(color(blue)(-4b-7))-36b = 6)" "larr# sub for #a#

#-108b-189-36b=6#

#-144b = 195#

#b = -195/144#

#a = -4(-195/144)-7#

#a = -19/12#