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Can you help me find the zeros step by step?

Can you help me find the zeros step by step?-example-1
User Inga
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1 Answer

3 votes

To find the zeros in a function you have to find those values that make that:


f(x)=0

In this case you have an cubic equation so first we are going to factorize it:


f(x)=x^3-x^2-5x+125^{}

Knowing that 125 is equal to:


125=5^3

this function can be expressed like:


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

The sum of cubes is:


a^3+b^3=(a+b)(a^2-ab+b^2)

We can use this to the first part of the equation:


(x^3+5^3)=(x+5)(x^2-5x+25)

Now we can factorize the other part as follow:


-x^2-5x=-x(x+5)

So the equation now is:


f(x)=\text{ }(x+5)(x^2-5x+25)-x(x+5)

Now we can factorize the (x+5) as a common term


(x+5)(x^2-5x+25-x)

And if we organice this one we get:


f(x)=(x+5)(x^2-6x+25)using he first part ( x + 5) we can find one zero, as follow:
x+5=0
x=-5

Finally we can use the quadratic equation in the second part we can find the other zeros, as follow:


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}
x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(25)}}{2(1)}
x=\frac{6\pm\sqrt[]{36-100}}{2}=\frac{6\pm\sqrt[]{-64}}{2}

As we get a root for a negative munber we can use the imaginary number:


\sqrt[]{-64}=8i

So:


x=(6\pm8i)/(2)

The zeros are now calculated with the two solutions ( + and -)


x_1=(6+8i)/(2)=3+4i
x_2=(6-8i)/(2)=3-4i

Then so, the zeros of the function


f(x)=x^3-x^2-5x+125^{}

are:

- 5

3 + 4i

3 - 4i

option B

User EMalik
by
7.9k points

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