# How does one know how to structure the unknown when factorizing using the remainder theorem?

When $x^3+x^2-7$ is divided by $(x-3)$, my book states that because the polynomial is cubic and the divisor is linear then the quotient must be quadratic and the remainder must be a constant. Basically, it sets put the following: $x^3+x^2-7=(Ax^2+Bx+C)\cdot(x-3)+D$

In another example, the book divides $x^4+x^3+x-10$ by $x^2+2x-3$, and states that because the divisor is quadratic and because the polynomial is quartic then quotient must be a quadratic and the remainder must be a linear expression.

I want to understand what logic they use here, in order to decide the morphology of the equation whose coefficients they will begin to find. I have been going at this question for a long time. No online teacher seems to tackle it, although it seems like a bog issue.

Another reason why I really want to understand this is because it arises in partial fractions, such as the partial fraction involved in the first order differential equation in question 7 of this booklet (https://madasmaths.com/archive/maths_booklets/further_topics/integration/1st_order_differential_equations_substitutions.pdf). The logic is even more baffling. Someone please help, I have a further maths exam coming up in 3 days and this isn't difficult, but just obscure.

[Note: please note that I don't want some complex mathemtical proof. I just want know the logic so that I can, with reasonable simplicity, be able to explain to someone else how to decide upon the morphology of the the right hand side when using the factor theorem and factorising a polynomial by inspection, in order to find the coefficients.]

• Are you familiar with what happens to the degrees when you multiply two polynomials? (It's easier to tackle multiplication before you get to division.) – Chris Culter Jun 2 '17 at 23:05
• Yes , of course I am. – Mathematician Jun 2 '17 at 23:30
• Why do you say "of course"? If you know what have to the degrees when you multiply two polynomials, what happens when you divide? – fleablood Jun 3 '17 at 0:18
• Recall the division algorithm for polynomials. $p(x)=q(x)\cdot g(x) + r(x)$ where $deg(r(x))=0$ or $deg(r(x))<deg(g(x))$. – Joe Jun 3 '17 at 0:29
• The logic is that the degree of the remainder is at most one less than the degree of tthe divisor. If it was equal or greater than the divisor you could divide further. Note: No-one ever said it was as high as one less. It's possible the polynomial will be a lower degree or be zero if it divides completely. In that case though it can still be technically be a polynomial of one degree less with leading zero coefficients. – fleablood Jun 3 '17 at 0:30

Lets talk about division algorithms, starting with real number and then relating it to polynomials

An integer is a polynomial of sorts.

That is: $1,603,279 = 1\times 10^6 + 6\times10^5 + 3\times 10^4 + 2\times 10^2+7\times 10 + 9$

and let say we want to divide this real number by $9.$

First, we know that the remainder will be less than 9. The remainder is always smaller than the divisor.

When we do the long division, we look for the largest number we can reasonably expect, that when multiplied by the divisor equals the number we want to divide.

Or, how many times does $9$ go into $16$...But we are secretly acknowledging that we are on the scale of $100,000.$

$1,603,279 = 900,000 + 1,603,278-900,000 = 900,000 + 703,279$

And now we say, how many times does $9$ go into $703,278?$

$1,603,279 = 9(170,000) + 703,278-630,000 = 9(170,000) + 73,279$

etc. until eventually we get to

$1,603,279 = 9(178,142) + 1$

You have been doing this since the 3rd grade, you just might not have thought about the process in this level of detail.

When we work with polynomials we are going to do the same thing.

$\frac {x^3+x^2-7}{x+3}$ What is the largest thing we can multiply by $x+3$ that will take the polynomial in the numerator down a degree?

how about $x^2$

$(x+3)(x^2) + x^3+x^2-7 - x^3 - 3x^2 = (x+3)(x^2) - 2x^2 - 7$

What is next to reduce the order of what remains.

$(x+3)(x^2 - 2x) - 2x^2 - 7 + 2x^2 + 6x = (x+3)(x^2 - 2x) +6x +7\\ (x+3)(x^2-2x + 6) + 6x+7 - 6x - 18 = (x+3)(x^2-2x + 6) -11$

$\frac {x^3+x^2-7}{x+3} = \frac {(x+3)(x^2-2x + 6) -11}{x+3} = x^2-2x + 6 - \frac {11}{x+3}$

The degree of the remainder will always be less than the degree of the divisor.

I hope this helps.

I just re-read the original post and feel like a little bit of an idiot, as I am explaining basic algebra, and you are looking to tackle diff eq's. Nonetheless, it is basic algebra.

• instructive to look at the other questions by the same OP, along with the comment threads after the answers... – Will Jagy Jun 3 '17 at 1:04
• @Will Jagy, once you learn through the british education system, you'll understand why I can do complex differential equations but have trouble grasping basic algebra... plus, I have 12 exams in 3 days, one after the other, so I tend to only skim the point atm. – Mathematician Jun 3 '17 at 4:22
• @Doug M, so is the general rule that the remainder will ALWAYS have a degree of one less than the largest degree of the divisor? I have always been taught by inspection and don't have time to learn a new method, but do you think I should always go ahead and put the remainder as having one degree less than the divisor? – Mathematician Jun 3 '17 at 4:45
• Not always, degree of one less, or smaller. – Doug M Jun 5 '17 at 16:21

I can't say that I understand anything about the question. However, one typically does not stop with a single step of the indicated type, $a = q b + r,$ where the degree of $q$ plus the degree of $b$ is the degree of $a.$ Usually this is part of the Euclidean algorithm. The degree of $r$ is, at most, smaller by $1$ that the degree of $b,$ but might turn out to be smaller. Note that constant polynomials have degree zero, while the zero polynomial is said to have undefined degree, or sometimes degree called $- \infty$

===========================================

$$\left( x^{3} + x^{2} - 7 \right)$$

$$\left( x - 3 \right)$$

$$\left( x^{3} + x^{2} - 7 \right) = \left( x - 3 \right) \cdot \color{magenta}{ \left( x^{2} + 4 x + 12 \right) } + \left( 29 \right)$$ $$\left( x - 3 \right) = \left( 29 \right) \cdot \color{magenta}{ \left( \frac{ x - 3 }{ 29 } \right) } + \left( 0 \right)$$ $$\frac{ 0}{1}$$ $$\frac{ 1}{0}$$ $$\color{magenta}{ \left( x^{2} + 4 x + 12 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} + 4 x + 12 \right) }{ \left( 1 \right) }$$ $$\color{magenta}{ \left( \frac{ x - 3 }{ 29 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + x^{2} - 7 }{ 29 } \right) }{ \left( \frac{ x - 3 }{ 29 } \right) }$$ $$\left( x^{3} + x^{2} - 7 \right) \left( \frac{ 1}{29 } \right) - \left( x - 3 \right) \left( \frac{ x^{2} + 4 x + 12 }{ 29 } \right) = \left( 1 \right)$$

====================================

$$\left( x^{4} + x^{3} + x - 10 \right)$$

$$\left( x^{2} + 2 x - 3 \right)$$

$$\left( x^{4} + x^{3} + x - 10 \right) = \left( x^{2} + 2 x - 3 \right) \cdot \color{magenta}{ \left( x^{2} - x + 5 \right) } + \left( - 12 x + 5 \right)$$ $$\left( x^{2} + 2 x - 3 \right) = \left( - 12 x + 5 \right) \cdot \color{magenta}{ \left( \frac{ - 12 x - 29 }{ 144 } \right) } + \left( \frac{ -287}{144 } \right)$$ $$\left( - 12 x + 5 \right) = \left( \frac{ -287}{144 } \right) \cdot \color{magenta}{ \left( \frac{ 1728 x - 720 }{ 287 } \right) } + \left( 0 \right)$$ $$\frac{ 0}{1}$$ $$\frac{ 1}{0}$$ $$\color{magenta}{ \left( x^{2} - x + 5 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} - x + 5 \right) }{ \left( 1 \right) }$$ $$\color{magenta}{ \left( \frac{ - 12 x - 29 }{ 144 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 12 x^{3} - 17 x^{2} - 31 x - 1 }{ 144 } \right) }{ \left( \frac{ - 12 x - 29 }{ 144 } \right) }$$ $$\color{magenta}{ \left( \frac{ 1728 x - 720 }{ 287 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 144 x^{4} - 144 x^{3} - 144 x + 1440 }{ 287 } \right) }{ \left( \frac{ - 144 x^{2} - 288 x + 432 }{ 287 } \right) }$$ $$\left( x^{4} + x^{3} + x - 10 \right) \left( \frac{ - 12 x - 29 }{ 287 } \right) - \left( x^{2} + 2 x - 3 \right) \left( \frac{ - 12 x^{3} - 17 x^{2} - 31 x - 1 }{ 287 } \right) = \left( 1 \right)$$

================================================

• p=x^3 +x^2-7=(x^2-9)+(x^3-27)+29=(x-3)(x^2+3x+9)+29 Therefore 29 must divide x-3 ; at least we must have: x-3=29 or x =32 and we get: 32^3+32^2-7=1165 . 29 That means if x = 29k+3 then p is divisible by x-3. – sirous Jun 3 '17 at 4:45

Work it through.

Suppose $$P(x) = \sum_{k=0}^n a_k x^k; a_n\ne 0$$ is an $$n$$ degree polynomial. And $$Q(x)= \sum_{k=0}^m b_k x^k; b_m\ne 0; m < n$$ is an $$m$$ degree polynomial.

To say $$P(x) = K(x)Q(x) + R(x)$$ with $$R(x)$$ having degree less than $$m$$ would be the same as saying $$P(x)- K(x)Q(x) = R(x)$$.

For that to happen $$K(x)Q(x)$$ must be a $$n$$ degree polynomial with all coefficients $$j: m\le j \le n$$ all equal to $$a_j$$.

We can claim

1. For that to happen $$K(x)$$ must be degree $$n-m$$.

Pf: If $$K(x)=\sum_{k=0}^w c_k x^k; c_w\ne 0$$ then $$K(x)Q(x) = c_wb_{m}x^{w+m} + (c_wb_{m-1} + c_{w-1}b_m x^{w+m+1} + ........ + (c_1b_0 + b_0c_1)x^1 + c_0b_0$$ will be a polynomial of degree $$w+m$$ with a leading coefficient of $$c_wb_m$$.

If $$w+m< n$$ then $$P(x) - K(x)Q(x)$$ will have a leading $$a_n$$ coefficient and will be a polynomial of degree $$n > m$$.

If $$w+m > n$$ then $$P(x) -K(x)Q(x)$$ will have a leading coefficient of $$-c_wb_m$$ and will be a polynomial of degree $$w+m > n >m$$.

So $$w+m =n$$ or $$w= n-m$$ is the only option.

But $$K(x)Q(x)$$ has leading of $$c_{n-m}b_m$$ so $$P(x)-K(x)C(x)$$ shouuld have a leading coefficient of $$a_n - c_{n-m}b_m$$ and if that is not equal $$0$$ then $$P(x) -K(x)C(x)$$ will have a degree of $$n$$.

So we can claim

1. If $$P(x) -K(x)Q(x)=R(x)$$ is a polynomial of degree less than $$n$$, then we must have $$c_{n-m} = \frac {a_n}{b_m}$$ and as $$b_m\ne 0$$ we can do this.

The next coefficient of $$K(x)Q(x)$$ is $$c_{n-m}b_{m-1} + c_{n-m-1}b_m$$ and the next coefficient of $$P(x) - K(x)Q(x) = a_{n-1} - (c_{n-m}b_{m-1} + c_{n-m-1}b_m)$$. If we want $$P(x)-K(x)Q(x)$$ to be a degree less than $$m \le n-1$$ we must have $$a_{n-1} - (c_{n-m}b_{m-1} + c_{n-m-1}b_m)=0$$ and therefore

1. $$c_{n-m-1} = \frac {a_{n-1} - c_{n-m}b_{m-1}}{b_m}$$ and we can do that as $$b_m \ne 0$$.

If $$n = m+1$$ were are done. $$K(x)Q(x) = a_nx^n + a_{n-1=m} x^{n-1=m} + \sum_{k=0}^{n-2}(\sum_{j=0}^kc_{n-m-k}b_k) x^k$$ so $$P(x) - K(x)Q(x)=R(x) = \sum_{k=0}^{n-2}(a_k -\sum_{j=0}^kc_{n-m-k}b_k) x^k$$, a uniquely determined polynomial of degree less than $$m$$.

However if $$n > m+1$$ we can keep repeating each step.

We will need $$a_{n-k} - (c_{n-m}b_{m-k} + c_{n-m-1}b_{m-k+1} + ...... + c_{n-m-k}b_m) = 0$$ and we have already determined what $$c_{n-m}, c_{n-m-1},....., c_{n-m-k + 1}$$ are.

1. That uniquely determines that we must have $$c_{n-m-k} = \frac {a_{n-k} - (c_{n-m}b_{m-k} + c_{n-m-1}b_{m-k+1} + ...... +c_{n-m-k+1}b_{m-1}}{b_m}$$.

2. And we do that until we unique determined what $$c_0$$ must be. And as $$K(x)$$ we uniquely determined $$R(x) = P(x) - K(x)Q(x)$$ is uniquely determined.

It might help to do and example to convince yourself:

Suppose we wanted to solve $$P(x) = x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 6x^3+7x^2+8x+ 9$$ and $$Q(x) = (4x^3 + 3x^2 + 2x + 1)$$. And we want $$P(x)= K(x)Q(x) + R(x)$$ where $$R(x)$$ has degree less than $$3$$.

Well, If $$K(x)$$ has degree $$w$$ then $$K(x)Q(x)$$ has degree $$3+w$$. And unless $$3+w=8$$ we will have a polynomial of degree $$8$$ minus a polynomial of degree $$3+w$$ will be of degree $$\max(8,3+w) > 3$$. So we must have $$w = 5$$.

So $$K(x) = (ax^5 + bx^4 + cx^3 + dx^2 + ex + f)$$.

And so $$K(x)Q(x) = (ax^5 + bx^4 + cx^3 + dx^2 + ex + f)(4x^3 + 3x^2 + 2x + 1) =$$

$$4ax^8 + (3a+4b)x^7 + (2a+3b+4c)x^6+(1a+2b+3c+4d)x^5 + (1b+2c + 3d +4e)x^4 + (1c + 2d+3e+4f)x^3 + (1d+2e + 3f)x^2 + (1e + 2f)x + f$$.

And for $$P(x) - K(x)Q(x)= R(x) =$$

$$(1 -4a)x^8 + (2 - (3a+4b))x^7 + (3- (2a+3b+4c))x^6+(4- (1a+2b+3c+4d))x^5 + (5- (1b+2c + 3d +4e))x^4 + (6-(1c + 2d+3e+4f))x^3 + (7- (1d+2e + 3f))x^2 + (8 - (1e + 2f))x + (9- f)$$

And if that is to have degree less than $$3$$ and we must have

$$1-4a =0$$

$$2-(3a+4b) = 0$$

$$3-(2a+3b+4c) =0$$

$$4- (1a+2b+3c+4d=0$$

$$5- (1b+2c + 3d +4e)=0$$

$$6-(1c + 2d+3e+4f)=0$$

So six linear equations and six unknowns. There is one uniqe set of solutions to that:

$$1-4a =0; a = \frac 14$$

$$2-(3a+4b) = 0; b=\frac {2-3a}4= \frac {2-\frac 34}{4}=\frac {\frac 54}{4}=\frac 5{16}$$

.... and so on......