## Saturday, September 22, 2012

### f(x) – f(x–1) = 3x^2 – 5x

After a lesson on the factor theorem, the teacher assigns the following homework problem.
“A cubic polynomial f(x) has a factor x and f(x) – f(x–1) = 3x2 – 5x.    …  Find f(x).”
(Question 28, page 97, New Additional Mathematics by Ho Soo thong and Khor Nyak Hiong)
Johnny begins as follows. As f(x) is cubic, he reasons that it will factorise as x(x – a)(x – b).
Consequently, Johnny defines f(x) in Wx Maxima. (A free computer algebra system that he downloaded from http://sourceforge.net/projects/wxmaxima/  .  Unfortunately, Johnny cannot use Wx Maxima on the exam.)

Note that (%i1) is “input 1” and (%o1) is “output 1”.

In a very straight-forward way, f(x) – f(x–1) is:

As the textbook has told Johnny that expression %o2 is identical to 3x2 – 5x, he inputs

Next he simplifies:

On comparing the coefficients of x Johnny sees that
–2b –2a  – 3 = –5
Thus, a =  1  –  b                                Equation (1)
As there is no constant term
(a + 1)b + a + 1 = 0     Equation (2)
Johnny substitutes for a:
(1  –  b + 1)b +  1  –  b +  1 = 0
b2 + b + 2 = 0
(– b + 2) (b  + 1) = 0
b = – 1 or b =2
To solve for a, he substitutes back into a = 1 – b.
If b =  – 1, a = 2  or if b = 2, a = –1.
Johnny realizes that it is immaterial which label goes with which number.  He has established that, written in factored form, f(x) = x(x –  –1) (x  – 2) = x(x + 1)(x – 2).

Alternatively, Mary starts with f(x) – f(x–1) = 3x2 – 5x and substitutes 0 for x, everywhere.
She gets f(0) – f(0 – 1) = 3(0)2 – 5(0) = 0
However she knows that f(0) itself evaluates to zero:
f(0) = 0(x – a) (x – b) =  ‘zero’ times ‘other stuff’ = 0.
Therefore,
0 – f(0 – 1) = 0
It follows that f(– 1) = 0
Mary applies the factor theorem, for any f(a) = 0 she knows (x –a) is a factor of f. Therefore (x –  – 1)
= (x + 1) is a factor of f.
Iteratively, Mary lets x = 1
f(1) – f(1–1) = 3(1)2 – 5(1)
f(1) – f(0) = –2
Once again she substitutes f(0) =0, so
f(1) – 0 = –2
f(1) = –2 ≠ 0, so (x – 1) is not a factor of f.
Therefore she continues and lets x = 2.
f(2) – f(2–1) = 3(2)2 – 5(2)
f(2) – f(1) = 3(2)2 – 5(2)
She substitutes for f(1).
f(2) – (–2) = 2
f(2) + 2 = 2
f(2) = 0, i.e., she knows (x – 2) is also a factor of f(x). Since f is a cubic function Mary knows it has exactly three factors.  She writes f(x) = x(x + 1)(x – 2).

Both students have successfully completed their homework.  Johnny has used the notion of function identity and successfully solved a pair of simultaneous equations to find his solution.  He made some use of the computer but could have expanded everything by hand if necessary. Mary has been more traditional and done all her calculations with pencil and paper. Also, she has directly applied the factor theorem itself, which is more in line with the teacher’s expectations.  Full marks to both students.

## Thursday, September 20, 2012

Q1.   3x+2  = 45 + 4 × 3x
Put all terms with 3x  on the left, 3x+2   – 4 × 3x    = 45
Factor out  3x  ,    3x  (3 – 4 )  = 45
…    Solve to get x = 2

Q2.     2 log5r – 3 log 125t = 2
Recall that 125 = 53, so try to get like terms on the left.  This means use the change of base formula.
2 log5r – 3 (log 5t / log 5125) = 2
2 log5r – 3 (log 5t / log 553) = 2
Next remember that log xx = 1, so log 55= 1
2 log5r – 3 (log 5t / log 553) = 2
Next the 3s will cancel
2 log5r – 3 (log 5t / 3) = 2
Replace 2 with log 5 25 and you are almost finished
…  r = 5t

Remember AB ≠ BA

So x = 3 and y = -5 .  (Confirm by substituting into both equations).

6.    4x+2 × 2y = 8
2, 4 and 8 are related by 4 = 22 and 8 = 23.
4x+2 × 2y = 8  à(22) x+2 × 2y = 23 à 2(x+2) + y = 3  Eqn (1)
Likewise, 27x – 2  × (1/3)y = 1 à (33) x – 2  × (3) – y = 30 à Eqn (2)
Don’t forget, x0 = 1   (unless 00)
… Solve for the simultaneous equations: x = 1, y =  – 3.