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Janice Mak (laputa@conneti.com) wrote:
: HERE, I GOT A MATH PROBLEM THAT I CAN'T SOLVE.
: IT'S FOR MY MATH 12 PROVINCIALS.
: 
: DETERMINE THE NUMBER OF REAL ROOTS OF THE EQUATION.
: 
: 4x(x-5)(x^2+9)

One of the more important results of Chapter 12 in the BC Math 12 text is:

   ROOTS  =  LINEAR FACTORS

Every linear factor, (x-a), of a polynomial gives you a root, x=a.
Every root, x=a, of a polynomial gives you a linear factor, (x-a).
(See the Factor Theorem on Page 50 of Addison-Wesley Mathematics 12.)

In your case, the only linear factors are (x-5) and x  [or (x-0)].
The 4 is a degree ZERO factor and doesn't give any roots.
The quadratic factor, (x^2+9), is irreducible (at least over the reals).
[Remember:  DIFFERENCE of Squares can be factored, SUMS of Squares not.]
So there are exacly two real roots, namely 5 and 0.

Followup:  Determine the number of real roots of the equation

           5x(x-6)(x-7)(x^2+4)(x^2-4)

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See http://www.bced.gov.bc.ca/exams/search/exsection.htm for final exam
practice questions with answers (some solutions).

>4x(x-5)(x^2+9)

has 2 real roots and 2 imaginary roots
the real roots are x=0, 5
imaginary roots are x= 3i, -3i