Sample question from a student, and my reply

Looking for feedback on whether I'm providing good help to my students via email. A sample (in fact, to this point, the only) email exchange for precalc.

How are you today? I am a student in your pre-calculus class.I have some questions on my homework.Could you please give me some help?

1.How to find out real zeros from all possible rational zeros of a function?(I have read the samples in the texrbook ,but I can't understand.Please give me a specific sample like page 160 :11)I know how to find all possible rational zeros ,but real zeros ,I can't.

2.Page 160:In exrcise 25-28 ,I have no idea to sketch the graph of f so that some of the possible zeros in part (a) can be disregarded)

Thank you for your help!Have a nice weekend.

My reply:

Hope you're enjoying your weekend so far!


1.How to find out real zeros from all possible rational zeros of a function?(I have read the samples in the texrbook ,but I can't understand.Please give me a specific sample like page 160 :11)I know how to find all possible rational zeros ,but real zeros ,I can't.

So the first thing to do is to use the rational zero test.

Remember that the rational zero test can be used on functions of the form

f(x) = an*x^n + a(n-1)*x^(n-1)+... + a1*x + a0.

where an is the coefficient of the term of order n (x^n).

We find the rational zeroes by taking the factors of the a0 term (the constant, which is -6) divided by the coefficient on the highest order term. In this case, the function is of order 3. The coefficient for x^3, a3, is 1.

If a0 is -6, the factors of a0 are {-6, -3, -2, -1, +1, +2, +3, +6}.

and the factors of a3 are {-1, 1}.

So, dividing the factors of a0 by the factors of a3 give us the possible rational zeroes:

a0/a3 = {-6, -3, -2, -1, 1, 2, 3, 6}.

I picked one root to try, +1. I substitute x = 1 into the function.

f(1) = (1)^3 - 6(1)^2 + 11(1) -6
= 0.

I got lucky! So I know that x = 1 is a rational root of f(x).

Now, there are two things I can do at this point.

1. Keep trying the other rational roots.

If you do this, you try the other possible rational roots (factors of a0 / factors of a3) and see if any of them are a zero.


2. Divide by x-1 to get a quadratic equation, which I will solve either by guessing or by using the quadratic formula.

Remember that if x = 1 is a root of f(x), (x - 1) is a factor of f(x), and we can divide f(x) by the factor to find the other roots.

This is the approach I used, because I didn't want to waste time in case there are no other rational roots. If x = 1 is the only rational root, then the quadratic formula will give us the real (or complex) roots, guranteed!

So, using long division,

(x^3 - 6x^2 + 11x - 6)/(x-1) = x^2 - 5x + 6


Now, I can guess the factors - it looks like x = 2 and x = 3 will work. Sure enough,

x^2 - 5x + 6 = (x - 2)*(x - 3)

which means the real roots of f(x) are x = {1, 2, 3}.

But it's a good idea to solve it using the quadratic formula, in case we can't factor it easily.

The quadratic formula can be used for any equation of the form

ax^2 + bx + c = 0

where a, b, and c are real numbers.

The solutions are found using this formula

roots = (- b +/- sqrt(b^2 - 4ac)/(2a)

where sqrt("something") equals the square root of "something".

For

x^2 - 5x + 6

a = 1, b = -5, and c = 6.

Using the formula,

roots = (- (-5) +/- sqrt((-5)^2 - 4*1*6))/(2*1)

= (5 +/- sqrt(25-24))/2

= (5 +/- 1)/2

= {2, 3}.

So we see that the solutions are x = 2 and x = 3.

In summary, for a polynomial function, you do the following steps:

1. Use the rational root test to find possible roots.

2. Try the roots and see if any of them work.

3. For every root that works, divide the function by its factor. For example, if the rational root is x = 4, you divide the function by the factor (x - 4).

4. After dividing, see if the function is of order 2 (looks like ax^2 + bx +c). If so, you can use the quadratic formula. This is how you get real roots that are irrational, and also complex roots.


2.Page 160:In exrcise 25-28 ,I have no idea to sketch the graph of f so that some of the possible zeros in part (a) can be disregarded)

The easiest way to graph it is to use your calculator. But if you don't have that, then the easiest way to graph is to find out what f(x) equals for a number of x values.

For example, on problem 25, f(x) = x^3 + x^2 - 4x - 4

Using the rational zero test, the possible rational roots are

x = {-4, -2, -1, 1, 2, 4}.

So I need to graph it at least over the domain of x:[-4, 4] (everywhere from x = -4 to x = 4).

I would write two columns like this

x f(x)

-4

-3

-2

-1

0

1

2

3

4

And calculate what f(x) is for each x.


x f(x)

-4 -36

-3 -10

-2 0

-1 0

0 -4

1 -6

2 0

3 20

4 60


If you draw the points and connect them, they look something like the picture I've attached in this email.

Note that from this picture, it's easy to eliminate -4, 1, and 4 as possible roots, but -2, 1, and 2 look like rational roots of f(x).

This is an easy example, but sometimes you'll get a function that is more complicated. That's why it's a good idea to graph it, especially in cases where you have a lot of possible rational roots to test.

Good luck! And thanks for asking good questions!