kitchen table math, the sequel: Algebra Review for Pre-Calculus (follow-up)

Thursday, July 28, 2011

Algebra Review for Pre-Calculus (follow-up)

I recently wrote about Algebra Review for Pre-Calculus and discussed some problems that I typically assign in the College Algebra courses I teach. Here is a link to some examples of those problems including algebraic simplification, simplifying rational expressions and solving equations involving rational expressions.

I'll write next about the Rational Roots Theorem and polynomial factorization. Here's a problem I found from a Japanese University Entrance Exam from 1990 that makes interesting use of these concepts. The document I took this from is here - the problem is from the first sample test on page 4.

Suppose the polynomial P(x) with integer coefficients satisfies the following conditions:

(A) If P(x) is divided by x^2-4x+3, the remainder is 65x-68

(B) If P(x) is divided by x^2+6x-7, the remainder is -5x+a

Then we know that a={?}.

Let us find the remainder bx+c when P(x) is divided by x^2+4x-21.
Condition (A) implies that {?}b+c={?}.
Condition (B) implies that {?}b+c={?}.
It follows that b={?} and c={?}

I've changed the notation of the answers a little in the hope of making it less confusing - you can see the original by clicking through the link.

I got pretty tangled up in solving this problem because I had never seen the Remainder Theorem used in quite this way before. The answers for this are on page 5 of the original.


Rich Beveridge said...

In using the Remainder Theorem to solve this you actually have to split the original divisors into their binomial factors. Notice that the two divisors share a factor.

GoogleMaster said...

I solved this by assuming that P(x) is quadratic, but I realize that's not necessarily a valid assumption, and I don't remember enough to know whether you could get a different answer from a higher order polynomial.

Dee Hodson said...

Hi Catherine
This should probably go on the main page but I still havent figured out how to do that...

In a recent conversation I was speaking about the high percentage of students entering universities that have to take remedial math. I think this point was mentioned in Waiting for Superman as well...

I have a disconnect between the SAT scores necessary to get admitted to these schools and the large percentage of kids who apparently cannot do college level math. How can kids score high on the math portion of the SAT yet still not be able to do math?

Am I missing something here? I would love any insights that the KTM community may provide..

Dee Hodson

Rich Beveridge said...

If you assume P(x)/(x^2-4x+3)=Q(x)+(65x-68)/(x^2-4x+3)

then multiply through by x-3, so


then the remainder when P(x) is divided by (x-1) is also 65x-68, so you can say that P(1)=65(1)-68=-3, or if you do the long division for (65x-68)/(x-1) you come out with a remainder of -3.

Do the same thing to condition (B) so that -3=-5(1)+a and a=2.

I don't know if there's an easier way to go about this or not...

Catherine Johnson said...

I have a disconnect between the SAT scores necessary to get admitted to these schools and the large percentage of kids who apparently cannot do college level math.

Hi Dee - Sorry you're having trouble posting - !

Do we know how high the remediation rate is at selective colleges?

Assuming high-math-SAT students admitted to selective colleges are taking remedial courses, the answer would be that SAT-I tests content from algebra 1, geometry (not including proofs), and just a bit of what I think of as algebra 2.

Catherine Johnson said...

Hey Rich - any chance you could post the solutions to your questions? (Or email them to me?)


Rich Beveridge said...

I have answers keys on my computer at work - I'll try to get them up this week.

ChemProf said...

Catherine's right, there is a disconnect between college level remedial math at selective colleges and what we mean when we talk about remedial math at a community college. A remedial math course at a private liberal arts college is probably at least algebra II (sometimes misleadingly called college algebra), which isn't tested on the math SAT.

At my moderately selective school, a chunk of students interested in STEM need to start with pre-calculus, which we consider remedial. A few could really use an algebra course, but they were usually admitted as transfers, who don't need SATs at all.