Question 718204: Given f(x)=x^3+2x^2-13x+10, (a) list all possible rational zeros, (b) find all rational zeros, and (c) factor f(x)
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Given:
(1) 
The first operation we need to do is FACTOR the trinomial! Then we can easily find the zeroes (roots). Our first step in the factoring procedure is to always remember that ALL odd power polynomials have AT LEAST one real root, ie a value of the variable, x in (1) that makes the polynomial equal to zero. Since x^3 is an odd power, (1) has one real root. How do we get it?
I start by setting the last two terms of the polynomial equal to zero and get
(2) -13x +10 = 0 or
(3) x = 10/13 which is close to 1, so let's see if when x = 1 in (1) we get zero.
(4) f(1) = 1 + 2*1 -13*1 +10 or
(5) f(1) = 3 + 10 -13 or
(6) f(1) = 0, hurrah! we got lucky and found a real root of (1)
Now what do we do?
Well, since x = 1 is a root of (1), (x-1) is a factor of (1). So now what we do is factor (x-1) out of (1). This is done by algebraic long division. I'm not able to show you this process here. Look in your text.
I determined that
(7) f(x) = (x-1)*(x^2+3x-10) which you can verify by multiplying to get
(8) f(x) = x^3 +3x^2 -10x -x^2 -3x +10 or
(9) f(x) = x^3 +2x^2 -13x +10 which is the same as (1)
Now we need to factor the quadratic of (7) and get
(10) (x^2+3x-10) = (x+5)*(x-2) which I assume you can do. If not see your text.
We now have factored the trinomial (1) as
(11) f(x) = (x-1)*(x+5)*(x-2)
All of the roots (zeroes) are
(12) x = (-5,1,2)
Answers:
(a) It's not possible to "list" an infinite set of numbers so we use,
zeroes = {x|x is all rational numbers}
(b) zeroes = (-5,1,2)
(c) f(x) = (x-1)*(x+5)*(x-2)
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