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My last session disconnected.Find rational zeros and other zeros. Solve f(x)=0. Then factor into linear factors: f(x)= x^3+3x^2-2x-6

User Grmihel
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Given the following function:


f\mleft(x\mright)=x^3+3x^2-2x-6

You can use the Rational Root Theorem to find the Rational zeros of the function. According to this theorem, if a polynomial has Rational zeros, eachey must have this form:


(p)/(q)

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

Then, in this case, you can identify that the Constant term is:


a_0=-6

And the Leading coefficient is:


a_n=1

Then, you can set up that:


(p)/(q)=(\pm1,2,3,6)/(\pm1)

So the possible zeros are:


=\pm1,\pm2,\pm3,\pm6

Now you must evaluate each case by substituting each value into the function, in order to find the actual Rational zeros:

For:


\begin{gathered} x=1 \\ x=-1 \\ x=2 \\ x=-2_{} \\ x=3 \\ x=-3 \\ x=6 \\ x=-6 \end{gathered}

You get:


\begin{gathered} f(1)=(1)^3+3(1)^2-2(1)-6=-4 \\ \\ f(-1)=(-1)^3+3(-1)^2-2(-1)-6=-2 \\ \\ f(2)=(2)^3+3(2)^2-2(2)-6=10 \\ \\ f(-2)=(-2)^3+3(-2)^2-2(-2)-6=2 \\ \\ f(3)=(3)^3+3(3)^2-2(3)-6=42 \\ \\ f(-3)=(-3)^3+3(-3)^2-2(-3)-6=0\text{ (It is a Rational root)} \\ \\ f(6)=(6)^3+3(6)^2-2(6)-6=306\text{ } \\ \\ f(-6)=(-6)^3+3(-6)^2-2(-6)-6=-102 \end{gathered}

Then, you know that this is a Rational zero of the function:


x=-3

To find the other zeros, you can follow these steps:

1. Divide the polynomial:


x^3+3x^2-2x-6

By this polynomial:


x+3

You can do it using Long Division:

Then, you can set up that:

The steps to solve it are shown below:

- Divide the first term of the Dividend by the first term of the Divisor.

- Place the result on the top.

- Multiply the result by the Divisor.

- Place the product below the Dividend.

- Subtract them in order to create a new polynomial.

- Bring down the next term and apply the same procedure.

Then, you get:

2. Therefore, you can rewrite the function as follows:


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

2.Make:


f(x)=0

3. And Set up this equation and solve for "x" in order to find the other roots:


\begin{gathered} 0=x^2-2 \\ 2=x^2 \\ \\ x=\pm\sqrt[]{2} \\ \\ x_1=\sqrt[]{2} \\ x_2=-\sqrt[]{2} \end{gathered}

Knowing those roots, you can factor the function into linear factors:


f\mleft(x\mright)=(x+3)(x+\sqrt[]{2})(x-\sqrt[]{2})

Therefore, the answers are:

- Rational zeros:


x=-3

- Other zeros:


\begin{gathered} x_1=\sqrt[]{2} \\ x_2=-\sqrt[]{2} \end{gathered}

- Factored into linear factors:


f\mleft(x\mright)=(x+3)(x+\sqrt[]{2})(x-\sqrt[]{2})

My last session disconnected.Find rational zeros and other zeros. Solve f(x)=0. Then-example-1
My last session disconnected.Find rational zeros and other zeros. Solve f(x)=0. Then-example-2
User William Le
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