Question 1158341


, the remainder is double that obtained when the expression is divided by (x+1). Show that c can have any value and find b in terms of a.

{{{ax^3 + bx^2 + cx - 4}}} is divided by {{{(x+2)}}}:

..........({{{ax^2+(b- 2a)x+(c-2b+4a)}}}
{{{(x+2)}}}|{{{ax^3 + bx^2 + cx - 4}}}
..........{{{ax^3 + 2ax^2}}} ............subtract
..................{{{(b- 2a)x^2+ cx}}}
.................{{{(b- 2a)x^2+2(b- 2a)x}}}..........subtract
............................{{{(c-(2b- 4a))x}}}
............................{{{(c-2b+4a)x- 4}}}
............................{{{(c-2b+4a)x+2(c-2b+4a)}}}.........subtract
.........................................{{{-4-2(c-2b+4a)}}}
.........................................{{{-4-2c+4b-8a}}}->reminder


{{{ax^3 + bx^2 + cx - 4}}} is divided by {{{(x+2)}}}:



..........({{{ax^2+(b- a)x+(c-b+a)}}}
{{{(x+1)}}}|{{{ax^3 + bx^2 + cx - 4}}}
............{{{ax^3 + ax^2}}} ............subtract
..................{{{(b- a)x^2+ cx}}}
.................{{{(b- a)x^2+(b- a)x}}}..........subtract
............................{{{(c-(b- a))x}}}
............................{{{(c-b+a)x- 4}}}
............................{{{(c-b+a)x+(c-b+a)}}}.........subtract
.........................................{{{-4-(c-b+a)}}}
.........................................{{{-4-c+b-a}}}->reminder


the remainder {{{-4-2c+4b-8a}}} is double more than remainder {{{-4-c+b-a}}}:

{{{-4-2c+4b-8a=2(-4-c+b-a)}}}

{{{-4-2c+4b-8a=-8-2c+2b-2a}}}........simplify

{{{-4-cross(2c)+4b-8a=-8-cross(2c)+2b-2a}}}-> so,this proves that {{{ c}}} can have {{{any }}}value

{{{-4+4b-8a=-8+2b-2a}}}.........solve for {{{b}}}

{{{4b-2b=8a-8+4-2a}}}

{{{2b=6a-4}}}

{{{b=3a-2}}}