Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the polynomial $f(x)=x^3+ax+b$, where $a$ and $b$ are constants. If $f(x+1004)$ leaves a remainder of $36$ upon division by $x+1005$, and $f(x+1005)$ leaves a remainder of $42$ upon division by $x+1004$, what is the value of $a+b$?

share|improve this question
    
The remainders shall hold for all $x$? –  Daniel Fischer Jul 29 '13 at 12:42
    
What have you tried as far as work? Did you attempt any computations with $\frac{f(x+1004)}{x+1005}$ and the other? –  Eleven-Eleven Jul 29 '13 at 12:43

3 Answers 3

up vote 0 down vote accepted

The general knowledge to be used is that if

$$P(x)=(x-a)Q(x)+R$$ where $P$ and $Q$ are polynomials and $R$ a constant. Then, evaluating on $x=a$ we get $P(a)=R$. This is called the Polynomial remainder theorem.

You are looking for $f(1)-1=a+b$.

We have that $f(x+1005)$ gives remainder $42$ after division by $x+1004=x-(-1004)$. This means that $f(-1004+1005)=f(1)=1+a+b=42$.

From this we get that $a+b=41$.

share|improve this answer

By assumption we have $$f(x+1004)=(x+1005)Q(x)+36\tag{1}$$ and $$f(x+1005)=(x+1004)S(x)+42\tag{2}$$ so let $x=-1005$ in $(1)$ gives $$f(-1)=-1-a+b=36$$ and let $x=-1004$ in $(2)$ gives $$f(1)=1+a+b=42$$ hence $$a=2\quad,\quad b=39$$

share|improve this answer
    
excellent, my friend! –  amWhy Apr 29 at 12:46

HINT: From the given information $$f(x+1004)=36+(x+1005)b_{1}(x)\\ f(x+1005)=42+(x+1004)b_{2}(x)$$ So, $$f(-1)=-a+b-1=36\\ f(1)=a+b+1=42$$

share|improve this answer
    
You have a typo in $f(-1)$: should be $-a+b-1=36$. –  Andrea Orta Jul 29 '13 at 12:50
    
Oh! sorry. I've edited now. –  Samrat Mukhopadhyay Jul 29 '13 at 12:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.