# Can the quotient of a division of a polynomial act as the residual in the remainder theorem?

The problem is as follows:

The polynomial $$p(x)$$ is divisible separately by $$(x-2)$$, $$(x-3)$$, $$(x-4)$$, and by $$(x-5)$$. If $$p(x)$$ is divided by $$(x-9)$$ leaves a remainder equal to $$5040$$, what is the independent term of such polynomial?

The alternatives given in my book are as follows:

$$\begin{array}{cc} 1.&\textrm{600}\\ 2.&\textrm{360}\\ 3.&\textrm{740}\\ 4.&\textrm{720}\\ \end{array}$$

The official solution for this problem according to my workbook in precalculus states the following argument:

$$p(x)$$ is divisible separately by $$(x-2)$$, $$(x-3)$$, $$(x-4)$$ and $$(x-5)$$ $$\Leftrightarrow$$ $$p(x)$$ is divisible by $$(x-2)$$, $$(x-3)$$, $$(x-4)$$ and $$(x-5)$$.

Using the division algorithm we have:

$$p(x)= (x-2)(x-3)(x-4)(x-5)\cdot q(x)$$

Therefore $$\operatorname{grad}[q(x)]=0$$ (comment: here the notation grad indicates the degree of the polynomial)

Hence $$q(x)=M$$

Using the remainder theorem it is obtained:

$$\frac{p(x)}{x-9}$$

$$p(9)=(9-2)(9-3)(9-4)(9-5)q(x)=5040$$

Then:

$$(7)(6)(5)(5)(4)M=5040$$

$$M=6$$

Then to get the independent term we do evaluate the polynomial in zero as: $$p(0)$$

$$p(0)=(0-2)(0-3)(0-4)(0-5)(6)=6\cdot 20 \cdot 6 =720$$

which corresponds with the option number $$5$$.

That's where it ends the official solution.

Now my problem is that I'm confused with the way how it was found the degree of the quotient and why it was earlier declared as such and later used as a remainder.

First and foremost, how do we know the degree of the quotient is zero? I'm only assuming that since this polynomial is divisible by those factors it implies an exact division, hence the quotient can only be a constant term, otherwise it will increase an additional factor for which the polynomial is not divisible with. But is this assumption correct?. Can someone explain this?

The second part for which I'm confused is a weird usage of the polynomial remainder theorem.

If my memory serves me right, the remainder theorem goes as this:

When a polynomial is divided by $$(x-a)$$ then

$$p(x)=d(x)\cdot q(x)+r$$

$$p(x)=(x-a)q(x)+r$$

When we do evaluate the polynomial in a thus:

$$p(a)=r$$

Hence evaluating the polynomial with that $$a$$ will result in the remainder of the division which it can be another polynomial, right? But when it is applied to a linear quotient then the remainder is a constant term.

But in this case, the way how it was made such interpretation is some confusing because it uses $$M$$ and $$M$$ was earlier declared as the quotient.

Shouldn't it be as:

$$p(x)=(x-9)((x-2)(x-3)(x-4)(x-5)\cdot q(x))+r=5040$$

Because of these inconsistencies. I'd appreciate someone could help me to understand what might the author had intended to do, because I'm lost. Perhaps does it exist a better method of argumentation here?. I appreciate if someone could explain with words the right interpretation of the polynomial remainder theorem because this is the part which I'm struggling the most. How is it being used here?.

The part which it confuses me a lot is why is it treating $$M$$ as if it were $$r(x)$$ if it was earlier declared as quotient?. Isn't this a contradiction when evaluates the polynomial on $$p(9)$$?. So all and all, can someone help me to address these doubts please?

• You are right, the first problem is: "Therefore $\operatorname{deg}(q)=0$." No, this is not true. We have $p=(x-2)(x-3)(x-4)(x-5)q$ with some polynomial $q$. This means, $p$ is divisible by $(x-n)$ for $n=2,3,4,5$. Secondly, $p=(x-9)g+5040$ for some polynomial $g$. Feb 12 at 10:32
• One thing why you might be confused is that you seem to mix two instances of division together. You wrote $p(x)=(x-9)((x-2)(x-3)(x-4)(x-5)\cdot q(x))+r=5040$, which is wrong for two reasons. Firstly, this means $p(x)=(x-9)p(x)+r$, which is not true -- the quotient is not $p(x)$ again, it is some different polynomial $t(x)$. Secondly, only $r=5040$, not the whole thing, when you leave $x$ as a variable. However, and this is IMO the main point of the exercise, if you substitute $9$ into $x$, the whole part $(x-9)t(x)$ disappears, and so you never really care what $t(x)$ is. Feb 12 at 11:42

For the first problem, as others pointed out, you are right that nothing guarantees $$\mathrm{deg}(q)=0$$ if we take the exercise as it is written. But perhaps what the author(s) meant is $$p(x)$$ is divisible by [...], and ONLY by those'', which would then imply that $$\mathrm{deg}(p)=4$$, and so necessarily $$\mathrm{deg}(q)=0$$.

Now for the second part. The reminder theorem says that for polynomials $$p(x), s(x)$$, there always exist polynomials $$q(x), r(x)$$ such that $$p(x) = s(x)\cdot q(x)+r(x),$$ where (!) $$\mathrm{deg}(r)<\mathrm{deg}(s)$$. Meaning that we can divide $$p$$ by $$s$$ as much as possible'', in the sense that whatever remains has degree strictly lower than the polynomial we are dividing by. In the exercise, we have two instances of division, and get two formulas: $$(1)\ \ p(x)=(x-2)(x-3)(x-4)(x-5)q(x),$$ $$(2)\ \ p(x)=(x-9)\cdot t(x)+r(x).$$ Formula $$(1)$$ just says that $$p$$ is divisible by all those linear factors, and $$q$$ is the quotient. Formula $$(2)$$ is the division of $$p$$ by $$(x-9)$$, where we get quotient $$t(x)$$ and some reminder $$r(x)$$, which actually has to be just a number, because $$\mathrm{deg}(r)<\mathrm{deg}(x-9)=1$$. Moreover, we are actually told that it is just a number, and even which number it is: $$r=5040$$.

Now we substitute $$x\rightarrow 9$$ into $$(2)$$, and get $$p(9)=(9-9)\cdot t(9)+r=r=5040.$$ But using $$(1)$$, this means that $$5040=p(9)=(9-2)(9-3)(9-4)(9-5)q(9)=7\cdot6\cdot5\cdot4\cdot q(9)=840q(9)$$ $$q(9)=5040/840=6$$

Now assuming that $$\mathrm{deg}(q)=0$$, we know that actually $$q=6$$ (the authors write this as $$M=6$$ just to stress out that $$q$$ is constant), and so $$p(x)=6\cdot(x-2)(x-3)(x-4)(x-5),$$ and it's easy to obtain the independent term. You can see it directly from the formula, that all you need to do is multiply $$6\cdot(-2)\cdot(-3)\cdot(-4)\cdot(-5)=720$$, but the authors use the general fact that $$p(0)$$ is the independent term for every polynomial, and so they evaluate $$p(0)$$, which is the same multiplication leading to $$720$$.

The only use of the remainder theorem is to recognise that $$p(9)=5040.$$

From the point where $$p(x)= M(x-2)(x-3)(x-4)(x-5)$$ simply substitute $$x=9$$ to obtain $$5040=7.6.5.4.M$$ i.e. $$M=6$$.

Multiplying out gives the constant term as $$2.3.4.5M=720$$.

P.S.

Your five lines saying what you remember of the remainder theorem are correct. Don't let the answer in the book confuse you!