Issues with "Elementary Algebra" by Hall and Knight

Question

I've started on Elementary Algebra by Hall and Knight about a week ago.

Things were going smoothly, but a technique called "Horner's Method of Synthetic Division," which is a variant of synthetic division, was introduced. However, I found the explanation of how to obtain the remainder to be unsatisfactory.

It shows the method of division as well as how to obtain the remainder. The problem with this is that I won't always know to how many terms I have to divide to in the division, so how do I just normally take the remainder? Because of this, I can not use the method of division provided by the text.

Then, after ignoring this and deciding to just use normal synthetic division as well as polynomial long division when necessary, I encountered another problem. The book explains how to obtain the HCF of algebraic expressions which cannot be easily factored. While the explanation was fine, when the same technique was applied in other examples, the explanation was inconsistent with the later applications of the method.

In the second use of the technique, why does is $3x^2-4x-4$ term used as a divisor twice? After this, I went on the internet to learn how to find the HCF of complex algebraic expressions and found out how; but why should I have to research how to do things that should be explained in the book?

I don't know if I am the problem, or whether the book is. I have very little of what is termed "mathematical maturity," so perhaps it is my incompetence that is creating issues. Or, the book is not very good.

Either way, I'd like for someone to perhaps clarify the inconsistencies, or tell me that I am the problem and am not reading the book correctly.

If someone has better book recommendations, that would be very helpful as well.

Thanks.

Answer 1

We want to find the Highest Common Factor (HCF) of

$$24x^4-2x^3-60x^2-32x = 2x(12x^3-x^2-30x-16 \\ 18x^4-6x^3-39x^2-18x = 3x(6x^3-2x^2-13x-6)$$

The factoring above already gives us an $x$ term, so we can ignore it for now.

We an compactly write this as a table (sorry I could not get the alignment correct)

$$\begin{array}{c|r|r}2x &6x^3-2x^2-13x-6 &12x^3-x^2-30x-16 &2\\&6x^3-8x^2-8x &12x^3-4x^2-26x-12 \\\hline 2&6x^2-5x-6&3x^2-4x-4&x\\&6x^2-8x-8 &3x^2+2x \\\hline &3x+2 &-6x-4 &-2 \\\hline &&-6x-4 \\\hline &&0\end{array}$$

The steps are

• We see that the $6x^3$ term goes into the $12x^3$ term two times, so we write a $2$ in the RHS and multiply the LHS by it and write that under, that is $=12x^3-4x^2-26x-12$ and subtract that from the item above $= 3x^2-4x-4$

• We now see that the $3x^2$ can go into the top LHS $6x^3$ term $2x$ times, so we write that in the top LHS, multiply that out $= 6x^3-8x^2 -8x$ and subtract from above $=6x^2-5x -6$

• We now see that the $3x^2$ term goes into the $6x^2$ term two times and write a $2$ in the LHS, multiply $=6x^2-8x-8$ and subtract from the upper term $=3x+2$

• We now see that the $3x$ term goes into the $3x^2$ term $x$ times, write that in the RHS, multiply $= 3x^2+2x$ and subtract from the upper term $=-6x-4$

• We now see that the $3x$ term goes into the $-6x$ term minus two times, write that in the RHS, multiply $=-6x-4$, subtract from the upper term and that is zero, so we are done.

• We keep doing this process until we can no longer do it.

The HCF is

$$x(3x+2)$$

Note: today programs like Mathematica (I am sure there is something in SAGE, Maple V and the like that can do this), we can write

   PolynomialGCD[24x^4-2x^3-60x^2-32x, 18x^4-6x^3-39x^2-18x]

This results in

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