SOLUTION: The function f is given by f(x)=x^3+2x^2+ax-8 where a is constant.when f(x) is divided by (x-2) the remainder is -6. Show that (x+1) is a factor of f(x)

Question 1159546: The function f is given by f(x)=x^3+2x^2+ax-8 where a is constant.when f(x) is divided by (x-2) the remainder is -6.
Show that (x+1) is a factor of f(x)
Found 3 solutions by ikleyn, MathLover1, MathTherapy:
Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.

According to the Remainder theorem, the fact that the remainder is -6, when f(x) is divided by (x-2), means that f(2) = -6. In other words, 2^3 + 2*2^2 + a*2 - 8 = -6. It implies 2a = -6 - 2^3 - 2*2^2 + 8 = -14. Hence, a = -14/2 = -7. Thus the polynomial f(x) is f(x) = x^3 + 2x^2 - 7x - 8. (1) Now, let us check that f(-1) = 0. For it, substitute x= -1 into the polynomial (1). You will get f(-1) = (-1)^3 + 2*(-1)^2 - 7*(-1) - 8 = -1 + 2 + 7 - 8 = 0. Now apply the Remainder theorem again. It says that if f(-1) = 0, then f(x) is divisible by the binomial (x-(-1)) = (x+1). It is EXACTLY what has to be proved. The proof is completed.

The problem is Solved.

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   Theorem   (the remainder theorem)
   1. The remainder of division the polynomial    by the binomial    is equal to the value    of the polynomial.
   2. The binomial    divides the polynomial    if and only if the value of    is the root of the polynomial , i.e. .
   3. The binomial    factors the polynomial    if and only if the value of    is the root of the polynomial , i.e. .

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    - Divisibility of polynomial f(x) by binomial x-a and the Remainder theorem
    - Solved problems on the Remainder thoerem
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"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".

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Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
given:
where is constant.when is divided by the remainder is .
first find :
.........(
|
.....................subtract
...............
...........................subtract
..............................
.......................
........................
.................................subtract
.........................................
.........................................
..................................................reminder
if given the remainder is , we have=>
then quotient is

and
=> since

subtract it from to find new reminder

=> reminder is
and

also given that:
can be written in the form whereand are constants
find the values of and
=>
.....expand right side

.....compare coefficients

substitute in

subtract

so, new reminder is , and

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
The function f is given by f(x)=x^3+2x^2+ax-8 where a is constant.when f(x) is divided by (x-2) the remainder is -6.
Show that (x+1) is a factor of f(x)

Factor: x - 2, so we get: x - 2 = 0___x = 2 <==== NON-ROOT

Factor: x + 1, so we get: x + 1 = 0___x = - 1 <==== One ROOT

As the above proves to be TRUE (0 = 0), x + 1 is DEFINITELY a factor of
I hope for your SANITY, you won't even look at the UNNECESSARY COMPLEX method presented by the other person!
Plus, the majority of what she has stated is WRONG, leading to WRONG answers, as USUAL!!
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