Question 1049745
Hello,I have a question on polynomials..I've spent much time trying to figure out what to do,but I keep getting answers that don't add up in the end..Here is the question: 
The polynomial f(x) is given by f(x) =4x^3 +ax^2 + bx+2. (4x+1) is a factor of f(x).When f(x) is divided by (x-1) there is a remainder of 20. Find a and b. 
Please help,I'm so stuck
---------------------------------------------------------------------------


The polynomial f(x) is given by f(x) ={{{4x^3 +ax^2 + bx+2}}}. {{{(4x+1)}}} is a factor of f(x).When f(x) is divided by {{{(x-1)}}} there is a remainder of {{{20}}}. Find {{{a}}} and {{{b}}}. 

solution
first step
when f(x) is evaluated at x={{{-1/4}}} the remainder is 0 Because {{{4x+1}}} is a factor of f(x).
f(x) ={{{4x^3 +ax^2 + bx+2}}}
x={{{-1/4}}}
{{{0}}} ={{{4*(-1/4)^3 +a*(-1/4)^2 + b*(-1/4)+2}}}

simplify to get
{{{a-4b}}}={{{-31}}}

second step
what happens when f(x) is evaluated at x={{{1}}}
f(x) ={{{4x^3 +ax^2 + bx+2}}}


{{{4+ a + b+2}}} this expression equals the remainder which we know to be {{{20}}}

{{{20}}}={{{4+ a + b+2}}}
{{{20}}}={{{6+ a + b}}}

{{{20-6}}}={{{ a + b}}}

{{{14}}}={{{ a + b}}}
{{{a+b}}}={{{ 14}}}



{{{a-4b}}}={{{-31}}}
{{{a+b}}}={{{ 14}}}


*[invoke linear_substitution "a", "b", 1, 1, 14, 1, -4, -31 ]


answer
{{{a}}}={{{5}}}
{{{b}}}={{{9}}}
the polynomial is

f(x) ={{{4x^3 +5x^2 + 9x+2}}}

check that

{{{(4x^3 +5x^2 + 9x+2)}}}/{{{(4x+1)}}} gives off a remainder of {{{0}}} meaning
{{{(4x+1)}}} is a factor of f(x).


when
{{{(4x^3 +5x^2 + 9x+2)}}}/{{{(x-1)}}} the remainder is {{{20}}} which is the same as evaluating the expression {{{(4x^3 +5x^2 + 9x+2)}}} at x={{{1}}}
{{{(4 +5 + 9+2)}}}
{{{(20)}}}