Question #33e19

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Answer 1 · 2017-03-17T15:27:22

The divisor is

#x^2-5x+6#

#=x^2-3x-2x+6#

#=x(x-3)-2(x-3)#

#=(x-3)(x-2)#
so it has two linear factors #(x-3) and (x-2)#

The polynomial of #x# to be divided is

#P(x)=ax^3-9x^2+bx+3a#
This is to be exactly divisible by the divisor containing two linear factors #(x-3) and (x-2)#.

So

#P(3)=0#

#=>a*3^3-9*3^2+b*3+3a=0#

#=>30a+3b=729#

#=>10a+b=243........(1)#

Again

#P(2)=0#

#=>a*2^3-9*2^2+b*2+3a=0#

#=>11a+2b=36........(2)#

Multiplying (1) by 2 and subtracting (2) from the product we get

#20a-11a=486-36#

#=>9a=450#

#=>a=450/9=50#

Inserting the value of a in equation (1)

#=>10xx50+b=243)#

#=>b=243-500=-257#

Answer 2 · 2017-03-17T17:30:26

#a=2,b=7#

If #ax^3-9x^2+bx+3a# is exactly divisible by #x^2-5x+6# then

#ax^3-9x^2+bx+3aequiv 0 mod x^2-5x+6#

then #exists (cx+d) | ax^3-9x^2+bx+3a = (cx+d)(x^2-5x+6)#

or

#(a-c)x^3+(5c-9-d)x^2+(5d-6c+b)x+3a-6d=0#

Solving

#{(a-c=0),(5c-9-d=0),(5d-6c+b=0),(3a-6d=0):}#

we have

#a=2,b=7,c=2,d=1#