Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Math Help?

State the possible rational zeros for each function. Then factor each and find all zeros.

f(x) = x^3 - 3x^2 - 9x - 5

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Since the problem statement is to "state the possible rational zeroes," they want not only the actual zeroes, but all candidates by applying the rational root theorem.

Step 1, factor the constant term.  The factors are 5 and 1 (and -5 and -1, for these purposes).

Step 2, factor the coefficient of the highest power of x.  That's just 1 and -1.

The candidate roots are all possible combinations of the factors from Step 1, divided by all possible combinations of factors from step 2.  In this case, it's

5/1, 5/-1, 1/1, 1/-1, -5/1, -5/-1, -1/1, -1/-1

You can see that there are duplicates in the list.  Another way to do it is just use all positive numbers, then add minus signs after computing all the quotients.  So we have

5, -5, 1, and -1 as candidate roots.

Try each one in the original polynomial, and if you get 0, then it's a root.  For example, if you substitute -1 for x, you get 0.  That means that (x - -1) is a root.  aka (x+1)

In the above, if you got 3 zeroes immediately, you'd be done, but they probably didn't make it that easy.  You'll have to divide the original polynomial by (x+1) using synthetic division, and you'll be left with a quadratic, which you can solve to find the other zeroes.

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• x³-3x²-9x-5=(x - 5) (x + 1)²

(x - 5) (x + 1)²=0

So

(x-5)=0 and x=5

or

(x+1)=0 and x=-1

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• (x - 5) (x + 1)^2 = f(x)

zeros 5 and - 1

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• hint:

f(-1)=0 ---> x+1 is a linear factor

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