Question 1159546
given:

{{{f(x)=x^3+2x^2+ax-8}}} where {{{a}}} is constant.when{{{ f(x)}}} is divided by{{{ (x-2)}}} the remainder is {{{-6}}}.

first find {{{a}}}:

.........({{{x^2+4x+(a+8)}}}
{{{(x-2)}}} |{{{x^3+2x^2+ax-8}}}
.........{{{x^3-2x^2}}}............subtract
...............{{{4x^2+ax}}}
...............{{{4x^2-8x}}}............subtract
.......................{{{ax+8x}}}.......
.......................{{{(a+8)x}}}
........................{{{(a+8)x-8}}}
.......................{{{(a+8)x-2(a+8)}}}..........subtract
..................................{{{-8+2(a+8)}}}.......
..................................{{{-8+2a+8}}}.......
.........................................{{{2a}}}.........reminder

if given the remainder is {{{-6}}}, we have{{{ 2a=-6 }}}=> {{{a=-6}}}

then quotient is {{{x^2+4x+(a+8)=x^2+4x-3+8=x^2+4x+5}}}


and {{{ f(x)=x^3+2x^2-3x-8 }}}

=> since {{{(x^2 + 4 x + 5)(x - 2)=x^3 + 2 x^2 - 3 x - 10}}}


subtract it from {{{x^3+2x^2-3x-8}}}  to find new reminder

{{{x^3+2x^2-3x-8-(x^3 + 2 x^2 - 3 x - 10)=x^3+2x^2-3x-8-x^3 -2 x^2 +3 x + 10=2}}}

=> reminder is {{{2}}}

and {{{x^3+2x^2-3x-8 =(x^2 + 4 x + 5)(x - 2)+2}}}


also given that:

{{{ f(x)}}} can be written in the form {{{(x+1)(x^2+bx+c)}}} where{{{ b }}}and {{{c}}} are constants

find the values of {{{b}}} and {{{c}}}

{{{ f(x)=x^3+2x^2-3x-8 }}}=>
{{{ x^3+2x^2-3x-8=(x+1)(x^2+bx+c) }}}.....expand right side

{{{ x^3+2x^2-3x-8=bx^2 + bx + cx + c + x^3 + x^2}}}
{{{ x^3+2x^2-3x-8=x^3+ x^2+ bx^2 + bx + cx + c  }}}

{{{ x^3+2x^2-3x-8=x^3+ (1+ b)x^2 + (b + c)x + c  }}}.....compare coefficients

{{{2=(1+ b)}}}
{{{b=2-1}}}
{{{b=1}}}

{{{(b + c)=-3}}}
{{{1+c=-3}}}
{{{c=-4}}}

substitute in

{{{x^3+ (1+ b)x^2 + (b + c)x + c  }}}

{{{ x^3+ (1+ 1)x^2 + (1 -4)x -4  }}}
{{{ x^3+ 2x^2 +  -3x -4  }}}

subtract 

{{{x^3 + 2 x^2 - 3 x - 8 - (x^3 + 2 x^2 - 3 x - 4)=x^3 + 2 x^2 - 3 x - 8- x^3 - 2 x^2 +3 x + 4=-4}}}

so, new reminder is {{{-4}}}, and

{{{x^3 + 2 x^2 - 3 x - 8 = (x^2 + x - 4) (x + 1)  -4}}}