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Identify all the rational zeros for f(x) = x3 + x2 - 21x – 45.

User Cconnell
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Given the following function:


f\mleft(x\mright)=x^3+x^2-21x-45

You need to apply the Rational Roots Test:

1. By definition, if the function has Whole coefficients, then its Rational roots have this form:


(p)/(q)

Where "p" represents of the factors of the Constant term and "q" represents all the factors of the Leading coefficient.

2. Identify that Leading coefficient. This is:


a_n=1

3. Identify that Constant term. In this case this is:


a_0=_{}-45

4. Find all the factors of the Leading coefficient (positive and negative):


q=\pm1

5. Find all the factors of the Constant term (positive and negative):


p=\pm1,\pm3,\pm5,\pm9,\pm15,\pm45

6. You can set up that:


(p)/(q)=(\pm1,\pm3,\pm5,\pm9,\pm15,\pm45)/(\pm1)

7. Find all the combinations


\pm(p)/(q)

You need to divide the numerator by each denominator. Since the denominator is 1:


=\pm1,\pm3,\pm5,\pm9,\pm15,\pm45

8. Since there are no duplicates, the next step is to substitute each value into the function. Remember that, If:


P(a)=0

then "a" is a root of the polynomial.

Then:


\begin{gathered} f(-1)=(-1)^3+(-1)^2-21(-1)-45=-24 \\ \\ \\ f(1)=(1)^3+(1)^2-21(1)-45=-64 \\ \\ \\ f(-3)=(-3)^3+(-3)^2-21(-3)-45=0\text{ (}-3\text{ is a root)} \\ \\ \\ f(3)=(3)^3+(3)^2-21(3)-45=-72 \\ \\ \\ f(-5)=(-5)^3+(-5)^2-21(-5)-45=-40 \\ \\ \\ f(5)=(5)^3+(5)^2-21(5)-45=0\text{ (}5\text{ is a root)} \\ \\ \\ f(-9)=(-9)^3+(-9)^2-21(-9)-45=-504 \\ \\ \\ f(9)=(9)^3+(9)^2-21(9)-45=576 \\ \\ \\ f(-15)=(-15)^3+(-15)^2-21(-15)-45=-2,880 \\ \\ \\ f(15)=(15)^3+(15)^2-21(15)-45=3,240 \\ \\ \\ f(-45)=(-45)^3+(-45)^2-21(-45)-45=-88,200 \\ \\ \\ f(45)=(45)^3+(45)^2-21(45)-45=92,160 \end{gathered}

The answer is:


\begin{gathered} x=-3 \\ \\ x=5 \end{gathered}

User Wayne Wang
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