Question 1135670


When a polynomial {{{f(x)}}} is divided by {{{x-3}}} the remainder is {{{-9 }}}

{{{x-3=0}}}=>{{{x=3}}}

=>{{{f(3)=-9}}}

and when divided by {{{2x-1}}} the remainder is {{{-6}}}

{{{(2x-1)=0}}}=>{{{x=1/2}}}

=>{{{f(1/2)=-6}}}


We want the remainder when {{{f(x)}}} is divided by {{{(x-3)(2x-1)}}}. 

So let {{{f(x) = q(x)(x-3)(2x-1) + r(x)}}}, where {{{q}}} is the quotient and {{{r}}} is the remainder. 

Being the remainder on division by a quadratic, {{{r}}} must be a linear polynomial, say 
{{{r = ax+b}}}.

So {{{f(x) = q(x)(x-3)(2x+1) + ax + b}}}

{{{f(3) = -9}}} means {{{3a+b = - 9}}}.....eq.1
{{{f(1/2) = -6 }}}means {{{(1/2)a + b = -6}}}, or {{{a + 2b = -12}}}.....eq.2

Solve the system 
{{{3a+b = - 9}}}.....eq.1
{{{a + 2b = -12}}}.....eq.2
----------------------------
{{{3a+b = - 9}}}.....eq.1.....solve for {{{b}}}

{{{b = - 9-3a}}}......substitute that into eq.2

{{{a + 2(- 9-3a) = -12}}}.....eq.2

{{{a + - 18-6a = -12}}}

 {{{- 18-5a = -12}}}

 {{{- 18+12= 5a}}}
 
{{{- 6= 5a }}}

{{{a=-6/5}}}

{{{b = - 9-3a}}}......substitute {{{a}}}
{{{b =- 9-3(-6/5)}}}

{{{b =- 9+18/5}}}

{{{b =- 45/5+18/5}}}

{{{b =- 27/5}}}

Conclusion: the remainder is 

{{{r = ax+b}}}....substitute {{{a}}} and {{{b}}}

{{{r = -(6/5)x- 27/5}}}