# find variable from polynomial when divisor and remainder is known

I am asked the following question:

When ${x^4 - 3x^3 + px - 5}$ is divided by ${x - 3}$ the remainder is $16$, find $p$.

So my way of solving this was to use synthetic long division to divide ${x^4 - 3x^3 + px - 5}$ by $3$ that would leave me with a remainder expression that would have $p$. $$\begin{matrix}3&1&0&-3&p&-5\\ &&3&9&18&54+3p\\ &1&3&6&18+p&49+3p \end{matrix} %3| 1 | 0 | -3 | p | -5 % | 3 | 9 | 18 | 54 + 3p % % | 1 | 3 | 6 | 18 + p | 49 + 3p$$

I get the remainder to be ${3p + 49}$

I would say then that ${3p + 49 = 16}$ and ${p = - 11} But the textbook gives an answer of$7$. Have I come to the wrong conclusion about how to work out${p}$? • The remainder should be$3p-5$as the first two terms is a multiple of$x-3$. You must have made some mistakes in your long division. Unfortunately it is not in a human readable format so I can't say where exactly is wrong. Oct 29, 2015 at 10:45 ## 4 Answers Hint: Note that$f(x)=(x-3)\cdot g(x)+r(x)$implies$f(3)=r(3)$. Notice, The polynomial$x^4-3x^3+px-5-16$i.e.$x^4-3x^3+px-21$is completely divisible by$(x-3)$hence,$(x-3)$is a factor of$x^4-3x^3+px-21$Hence,$x=3$will satisfy the polynomial$x^4-3x^3+px-21$as follows $$(3)^4-3(3)^3+p(3)-21=0$$ $$3p=21\implies \color{red}{p=7}$$ • why is it - 16 and not 16 for the remainder that is added onto the polynomial? Oct 29, 2015 at 13:48 • OK, the remainder is always non-negative. Oct 29, 2015 at 14:19 • If I understand the remainder theorem correctly, it states:${f(x) = (x-c)·q(x) + r}$But in the answers here you are subtracting 16 rather than adding 16. I don't understand why Oct 29, 2015 at 20:03 • Notice, remainder is non-negative which is always subtracted from a given number for complete divisibility. It should not be added to the number as it will not make the number completely divisible Oct 29, 2015 at 20:06 Your question implies that$(x-3)|(x^4-3x^3+px-5-[16])\$ or, $$(x-3)|(x^4-3x^3+px-21)$$

Now say that $$x^4-3x^3+px-21=(x-3)(ax^3+bx^2+cx+d)$$ $$=ax^4+(b-3a)x^3+(c-3b)x^2+(d-3c)x-3d$$

So you get that easily comparing LHS and RHS that $$d=7,a=1,b=0,c=0$$ Hence $$p=d-3c=7$$

• Though your answer is right, you seem to have complicated things unnecessariy. Working with p(3) = 16 would have solved it instantaneously. I will not downvote it, since yours is also an approach to follow, and let this answer be very much here. Oct 29, 2015 at 11:07

Let $$P(x)= x^4-3x^3+px-5, \quad g(x)=x-3,$$ so, \begin{align} P(3)&= 3^4-(3)(3)^3+p(3)-5\\ P(3)&= 81-81+3p-5\\ P(3)&= 3p-5. \end{align} Note that the remainder is $$16$$. Hence, \begin{align} P(3) &= 16\\ 3p-5 &= 16\\ 3p/3 &= 21/3. \end{align} So, $$p = 7.$$

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