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)

Algebra ->  Permutations -> 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)      Log On


   



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) About Me  (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  f%28x%29  by the binomial  x-a  is equal to the value  f%28a%29  of the polynomial.
   2. The binomial  x-a  divides the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.
   3. The binomial  x-a  factors the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.


See the lessons
    - Divisibility of polynomial f(x) by binomial x-a and the Remainder theorem
    - Solved problems on the Remainder thoerem
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
given:
f%28x%29=x%5E3%2B2x%5E2%2Bax-8 where a is constant.when+f%28x%29 is divided by+%28x-2%29 the remainder is -6.
first find a:
.........(x%5E2%2B4x%2B%28a%2B8%29
%28x-2%29 |x%5E3%2B2x%5E2%2Bax-8
.........x%5E3-2x%5E2............subtract
...............4x%5E2%2Bax
...............4x%5E2-8x............subtract
.......................ax%2B8x.......
.......................%28a%2B8%29x
........................%28a%2B8%29x-8
.......................%28a%2B8%29x-2%28a%2B8%29..........subtract
..................................-8%2B2%28a%2B8%29.......
..................................-8%2B2a%2B8.......
.........................................2a.........reminder
if given the remainder is -6, we have+2a=-6+=> a=-6
then quotient is x%5E2%2B4x%2B%28a%2B8%29=x%5E2%2B4x-3%2B8=x%5E2%2B4x%2B5

and +f%28x%29=x%5E3%2B2x%5E2-3x-8+
=> since %28x%5E2+%2B+4+x+%2B+5%29%28x+-+2%29=x%5E3+%2B+2+x%5E2+-+3+x+-+10

subtract it from x%5E3%2B2x%5E2-3x-8 to find new reminder

=> reminder is 2
and x%5E3%2B2x%5E2-3x-8+=%28x%5E2+%2B+4+x+%2B+5%29%28x+-+2%29%2B2

also given that:
+f%28x%29 can be written in the form %28x%2B1%29%28x%5E2%2Bbx%2Bc%29 where+b+and c are constants
find the values of b and c
+f%28x%29=x%5E3%2B2x%5E2-3x-8+=>
+x%5E3%2B2x%5E2-3x-8=%28x%2B1%29%28x%5E2%2Bbx%2Bc%29+.....expand right side
+x%5E3%2B2x%5E2-3x-8=bx%5E2+%2B+bx+%2B+cx+%2B+c+%2B+x%5E3+%2B+x%5E2
+x%5E3%2B2x%5E2-3x-8=x%5E3%2B+x%5E2%2B+bx%5E2+%2B+bx+%2B+cx+%2B+c++
+x%5E3%2B2x%5E2-3x-8=x%5E3%2B+%281%2B+b%29x%5E2+%2B+%28b+%2B+c%29x+%2B+c++.....compare coefficients
2=%281%2B+b%29
b=2-1
b=1
%28b+%2B+c%29=-3
1%2Bc=-3
c=-4
substitute in
x%5E3%2B+%281%2B+b%29x%5E2+%2B+%28b+%2B+c%29x+%2B+c++
+x%5E3%2B+%281%2B+1%29x%5E2+%2B+%281+-4%29x+-4++
+x%5E3%2B+2x%5E2+%2B++-3x+-4++
subtract

so, new reminder is -4, and
x%5E3+%2B+2+x%5E2+-+3+x+-+8+=+%28x%5E2+%2B+x+-+4%29+%28x+%2B+1%29++-4

Answer by MathTherapy(10551) About Me  (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 matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+x%5E3+%2B+2x%5E2+-+7x+-+8%29
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!!