SOLUTION: If x+2 is a factor of x^3 - ax- 6,then find the remainder when 2x^3+ax^2-6x+9 is divided by x+1. Please help me with this question. Thank You.

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Question 1088858: If x+2 is a factor of x^3 - ax- 6,then find the remainder when 2x^3+ax^2-6x+9 is divided by x+1.
Please help me with this question.
Thank You.

Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A root of x%5E3-ax-6 is -2.
-2    |   1   0   -a     -6
      |
      |      -2   4    -8+2a
      |_____________________________
          1   -2   4-a   2a-14

The remainder must be zero, 0.
cross%28a-14=0%29
2a-14=0
a-7=0
highlight_green%28a=7%29.

Now you want the synthetic division of 2x%5E3%2B7x%5E2-6x%2B9 divided by x%2B1.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
If x+2 is a factor of x^3 - ax- 6,then find the remainder when 2x^3+ax^2-6x+9 is divided by x+1.
Please help me with this question.
Thank You.
f%28-+2%29+=+x%5E3+-+ax+-+6
Solving this by substituting 0 for f(- 2), and - 2 for x, you will get a = 7.
Since x + 1 is a factor, then a zero is - 1. You now need to use either POLYNOMIAL LONG DIVISION to DIVIDE 2x%5E3+%2B+7x%5E2+-+6x+%2B+9 by x + 1, or synthetic division to determine the remainder.
IGNORE the rubbish the other person POSTED. You'll only CONFUSE yourself a great deal if you decide to follow him. Most people already know this. Be FOREWARNED!!

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
If x+2 is a factor of x^3 - ax- 6,then find the remainder when 2x^3+ax^2-6x+9 is divided by x+1.
Please help me with this question.
Thank You.
~~~~~~~~~~~~~~~~~

Let me do it in a way in how it SHOULD BE DONE.

1.  If (x+2) is a factor of x%5E3+-ax+-+6, then, according to the Remainder theorem, the number -2 is the root of the polynomial p(x) = x%5E3+-ax+-+6.

    In other words, p(-2) = 0,  which means

    %28-2%29%5E3-a%2A%28-2%29+-+6 = 0,   or, simplifying,

    -8 + 2a - 6 = 0,   or   2a = 8 + 6 = 14  ====>  a = 14%2F2 = 7.



3.  Hence, the second polynomial is 

    q(x) = 2x%5E3+%2B+7x%5E2+-+6x+%2B+9.


    Then,  again, due to the Remainder theorem, the remainder of division q(x) by (x+1) is equal to the value q(-1), i.e.

     q(-1) = 2%2A%28-1%29%5E3+%2B+7%2A%28-1%29%5E2+-+6%2A%28-1%29+%2B+9 = 2*(-1) + 7*1 + 6 + 9 = 20.


Answer. The remainder of division the second polynomial by (x+1) is 20.

Solved.


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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 lesson
    - Divisibility of polynomial f(x) by binomial (x-a) and the Remainder theorem
in this site.

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

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


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Ignore writing by @josgarithmetic. His way IS NOT the method for solving such problems.

It is not his level of knowledge and is not his area of expertise.