Ray's question at Yahoo Answers regarding polynomial division

[MarkFL]

Here is the question:

Find R ,If R when p(x) is divided by..?


If R when p(x) is divided by (x-1) is -7 and R when p(x) is divided by (x+3) is 2,

find R when p(x) is divided by (x-1) and (x+3)..

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Ray,

I will show you two methods to solve this problem.

i) By the division algorithm, we may state:

\(\displaystyle P(x)=(x-1)(x+3)Q(x)+R(x)\)

We know the degree of the remainder must be at most one less than the divisor, and so we may state:

(1) \(\displaystyle P(x)=(x-1)(x+3)Q(x)+ax+b\)

And from the remainder theorem we know:

\(\displaystyle P(1)=-7\)

\(\displaystyle P(-3)=2\)

Using these data points with (1), we obtain:

\(\displaystyle P(1)=a+b=-7\)

\(\displaystyle P(-3)=-3a+b=2\)

Subtracting the second equation from the first, we eliminate $b$ and obtain:

\(\displaystyle 4a=-9\)

\(\displaystyle a=-\frac{9}{4}\)

Substituting for $a$ into the first equation, we find:

\(\displaystyle -\frac{9}{4}+b=-7\)

\(\displaystyle b=-\frac{19}{4}\)

And so we find the remainder is:

\(\displaystyle R(x)=-\frac{9}{4}x-\frac{19}{4}\)

ii) Here's an alternate approach:

\(\displaystyle \frac{P(x)}{x-1}=Q_2(x)+\frac{-7}{x-1}\)

\(\displaystyle \frac{P(x)}{x+3}=Q_1(x)+\frac{2}{x+3}\)

Subtract the second equation from the first:

\(\displaystyle P(x)\left(\frac{1}{x-1}-\frac{1}{x+3} \right)=\left(Q_2(x)-Q_1(x) \right)+\left(\frac{-7}{x-1}-\frac{2}{x+3} \right)\)

Simplifying by combining terms, and using the definition:

\(\displaystyle 4Q(x)\equiv Q_2(x)-Q_1(x)\)

we have:

\(\displaystyle \frac{4P(x)}{(x-1)(x+3)}=4Q(x)+\frac{-9x-19}{(x-1)(x+3)}\)

Dividing through by $4$, we obtain:

\(\displaystyle \frac{P(x)}{(x-1)(x+3)}=Q(x)+\frac{-\dfrac{9}{4}x-\dfrac{19}{4}}{(x-1)(x+3)}\)

And so we find the remainder is:

\(\displaystyle R(x)=-\frac{9}{4}x-\frac{19}{4}\)