Use the rational zeros theorem to find all the real zeros of the polynomial function. Use the zeros to factor f over the real numbers.

`f(x) = x^3 - 2x^2 - 13x - 10`

Find the real zeros of f. Select the correct choice below and, if necessary, fill in the answer box to complete your answer.

a. x=

or

b. There are no real zeros

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`f(-1)=0` so that: `x_1=-1`

`(x+1)(ax^2+bx+c)=x^3-2x^2-13x-10`

`ax^3+bx^2+cx+ax^2+bx+c=` `ax^3+x^2(a+b)+x(b+c)+c=x^3-2x^2-13x-10`

that gives:

`a=1,b=-3,c=-10`

so that we can write:

`x^3-2x^2-13x-10=(x+1)(x^2-3x-10)`

So: concerning solutions : `x^2-3x-10` :

`Delta= 9-4(-10)=49 >0` two real solution:

`x=(3+-sqrt(49))/2=(3+-7)/2` `x_2=5` `x_3=-2`

Finally the solution given equation are:

`x_1=-1; x_2=5;x_3=-2`

Graphicof cubic show the roots: `x_1=-1;x_2=5;x_3=-2`

Since `f(x) = x^3-2x^2-13x-10`

we can evaluate the function at all factors of `pm 10` to find possible zeros of the function. In this case, we see that

`f(-1)=-1-2+13-10=0`

which means that `x+1` is a factor of `f(x)` . Now use division to get the remaining factors, which gives:

`f(x)=(x+1)(x^2-3x-10)`

But the quadratic factor will further factor into two linear terms to get

`f(x)=(x+1)(x+2)(x-5)`

**This means that the zeros of the cubic are the negatives of the linear factors to get `x=-1` , `x=-2` and `x=5` .**

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