SOLUTION: a Polynomial P(x) and a divisor d(x) are given. Use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x), and express P(x) in the form d(x)*Q

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: a Polynomial P(x) and a divisor d(x) are given. Use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x), and express P(x) in the form d(x)*Q      Log On


   



Question 882930: a Polynomial P(x) and a divisor d(x) are given. Use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x), and express P(x) in the form d(x)*Q(x)+R(x)
P%28x%29=+x%5E3-x%5E2%2B7
d%28x%29=+x%2B2

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE SOLUTION:
highlight%28Q%28x%29=x%5E2-3x%2B6%29 , highlight%28R%28x%29=-5%29 , and P(x) in the form d(x)*Q(x)+R(x), or
P%28x%29=d%28x%29%2AQ%28x%29%2BR%28x%29 , is highlight%28P%28x%29=%28x%2B2%29%28x%5E2-3x%2B6%29-5%29 .

THE LONG DIVISION:
The format for long division varies from country to country (and maybe even from teacher to teacher. I will use the format my children were taught in the USA.
The first term of quotient Q%28x%29 is the quotient of the first terms x%5E3%2Fx=red%28x%5E2%29 .
Then red%28x%5E2%29%2Ad%28x%29=red%28x%5E2%29%28x%2B2%29=x%5E2%2B2x%5E2 is subtracted from P%28x%29=+x%5E3-x%5E2%2B7 .
Subtracting x%5E2%2B2x%5E2 is adding %28-1%29%28x%5E2%2B2x%5E2%29=-x%5E2-2x%5E2 ,
and the result is
P%28x%29%2B%28-x%5E2-2x%5E2%29=x%5E3-x%5E2%2B7-x%5E2-2x%5E2=-3x%5E2%2B7 .
That is shown below, but I am saving room for terms in x by including the term %220%22x .
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We continue dividing by x%2B2 the remaining -3x%5E2%2B7 ,
which I wrote as -3x%5E2%2B0x%2B7 to save room for the term in x .
The next term in the quotient, which comes from dividing first terms, is
%28-3x%5E2%29%2Fx=red%28-3x%29 .
That term, times the divisor is
%28red%28-3x%29%29%2Ad%28x%29=%28red%28-3x%29%29%28x%2B2%29=-3x%5E2-6x , which must be subtracted from
-3x%5E2%2B7=-3x%5E2%2B0x%2B7 .
Subtracting -3x%5E2-6x is adding %28-1%29%28-3x%5E2-6x%29=3x%5E2%2B6x ,
and the result is -3x%5E2%2B7%2B3x%5E2%2B6x=6x%2B7 .
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To find the next term of Q%28x%29,
we have to divide the first term of the remaining 6x%2B7
by the first term of divisor d%28x%29=x%2B2 :
6x%2Fx=red%286%29 .
So the last term of the quotient will be %22%2B%22red%286%29 ,
and we need to subtract the product
red%286%29%2Ad%28x%29=red%2A%286%29%2A%28x%2B2%29=6x%2B12 from the remaining 6x%2B7 .
Since subtracting 6x%2B12 is adding %28-1%29%286x%2B12%29=-6x-12 ,
I add -6x-12 to that remaining 6x%2B7 and find the remainder:
R%28x%29=6x%2B7%2B%28-6x-12%29=6x%2B7%2B-6x-12=highlight%28red%28-5%29%29 :

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