Question 1198421: For the polynomial function, (a) list all possible rational zeros, (b) find all rational zeros, and (c) factor f(x).
f(x)=x^(3)+x^(2)-37x+35
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Since the leading coefficient is 1, this means the possible rational roots of the polynomial is simply the factors of the last term, aka constant term (35).
Factors of 35 are: 1, 5, 7, 35
We'll also consider the negative version of each factor.
Possible rational roots: 1, -1, 5, -5, 7, -7, 35, -35
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Part (b)
Use the results of part (a) to check each possible rational root to see if it's an actual rational root.
The goal is to have f(x) = 0 for the given input x value.
Let's see if x = 1 is a root
f(x)=x^3+x^2-37x+35
f(1)=(1)^3+(1)^2-37(1)+35
f(1)=0
The result of 0 tells us that x = 1 is indeed a root or x intercept.
This is one location where the graph either touches the x axis (and bounces away) or crosses the x axis.
Try x = -1
f(x)=x^3+x^2-37x+35
f(-1)=(-1)^3+(-1)^2-37(-1)+35
f(-1)=72
The nonzero result tells us that x = -1 is not a root of f(x).
I'll let you try the other potential rational roots.
There are two others.
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Part (c)
The results of part (b) will play a key role in determining the factors.
For instance, we know that x = 1 is a root from the work done in part (b).
Subtract 1 from both sides to end up with x-1 = 0 to show that (x-1) is a factor.
In general, if x = p is a root, then (x-p) is a factor.
Therefore,
f(x) = x^3+x^2-37x+35
f(x) = (x-1)(x-p)(x-q)
where p & q are yet to be determined.
I'll let you find these values.
They are the other two roots I mentioned at the end of part (b).
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