Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder

Question

Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder,

Hey I would just like to know how to solve this as the question had me confused I don’t know if I am suppose to do long division on write it out like in the question, thanks

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Oct 29 '19 at 07:50
Yes, long division is the way to go. Do you know how to do it? — Matti P., Oct 29 '19 at 07:54
When it gives me the values yes, but I can’t see what to use as the dividend — Jeff Wiggle, Oct 29 '19 at 08:04
@JeffWiggle When you divide $p(x)$ by $x+1$, you get $Q(x)$ with a remainder of $R$. — Arthur, Oct 29 '19 at 08:04

score 0 · Answer 1 · answered Oct 29 '19 at 08:23

Any textbook would tell you to do polynomial division. Personally, I think polynomial division is too opaque of a technique to use before you know how to solve a question like this without it. So here is an example using what basically amounts to long division, but much less mysterious.

Take $p(x) = x^3+2x^2-3x+5$. We want to write this of the form $(x+1)Q(x) + R$ for some polynomial $Q$, and some polynomial $R$ with smaller degree than $x+1$ (which is to say, $R$ is a constant). We do this by building $Q$ term by term, reducing the degree of the remainder until we have our constant $R$.

The fact that the highest degree term of $p$ is $x^3$ means that the highest degree term of $Q$ must be $x^2$. So we add $(x+1)x^2$, and simultaneously subtract $x^3+x^2$ from the remainder. This doesn't change the value, but it lets us start on building our $Q$: $$ p(x) = (x+1)x^2 - (x^3+x^2) + x^3+2x^2-3x+5\\ = (x+1)x^2 + x^2-3x+5 $$ We have now successfully gone from a third degree remainder (all of $p$) to a second degree remainder. We are one step closer to a constant remainder. Next step: The highest degree term of the remainder is $x^2$, so the next term to add to $Q$ must be $x$: $$ p(x) = (x+1)x^2 - (x^2 + x) + (x+1)x + x^2-3x+5\\ = (x+1)(x^2+x) -4x+5 $$ And finally, in the last step, we see that the leading term of the current remainder is $-4x$, so the final term to add to $Q$ must be $-4$: $$ p(x) = (x+1)(x^2+x) + (x+1)(-4) - (-4x-4) -4x+5\\ = (x+1)(x^2+x-4) + 9 $$ And there we have it: $Q(x) = x^2+x-4$ and $R = 9$. If you do the long division, you might recognize how each step in the long division corresponds to something I did here, and vice versa, but long division is something of a mystery if one doesn't know that this is the process that lies behind it.

Also, I have been extra verbose here. If I were to give this problem on a test, I wouldn't necessarily expect more from an answer than, say,

$$ p(x) = x^3+2x^2-3x+5\\ = (x+1)x^2 + x^2-3x+5\\ = (x+1)(x^2+x) -4x+5\\ = (x+1)(x^2+x-4) + 9 $$

All-in-all more transparent and easier to follow than polynomial long division.

score 0 · Answer 2 · answered Oct 29 '19 at 18:57

Put $\,a = -1\,$ in $\ p(x)\, =\, (x-a)\,\dfrac{p(x)-p(a)}{\ x\,-\,a} + p(a),\,$ the Polynomial Remainder Theorem.

The quotient is exact by the Polynomial Factor Theorem, and is computable by the Polynomial Division Algorithm (which also computes the remainder).

Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder

2 Answers2