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Find all zeros of f(x)=x^3-5x^2+x-5​

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Answer:

x = 5 and x = -5

Explanation:

To find the zeros of the polynomial function f(x) = x^3 - 5x^2 + x - 5, you need to find the values of x for which f(x) equals zero:

f(x) = x^3 - 5x^2 + x - 5 = 0

Unfortunately, this polynomial is not easy to factor directly. You can use numerical methods or a calculator to approximate the roots. One commonly used method is the Rational Root Theorem.

The Rational Root Theorem states that if a rational number p/q is a root of the polynomial, where p is a factor of the constant term (in this case, 5) and q is a factor of the leading coefficient (in this case, 1), then p/q is a potential root.

In this case, the potential rational roots are ±1 and ±5, because the factors of 5 are ±1 and ±5, and the factors of 1 are ±1.

Now, you can try these values in the function to see if they result in f(x) = 0:

f(1) = (1)^3 - 5(1)^2 + 1 - 5 = 1 - 5 + 1 - 5 = -8 (not a root)

f(-1) = (-1)^3 - 5(-1)^2 - 1 - 5 = -1 - 5 - 1 - 5 = -12 (not a root)

f(5) = (5)^3 - 5(5)^2 + 5 - 5 = 125 - 125 + 5 - 5 = 0 (root)

f(-5) = (-5)^3 - 5(-5)^2 - 5 - 5 = -125 - 125 - 5 - 5 = -260 (not a root)

So, the zeros of the function f(x) are x = 5 and x = -5. The polynomial has two real roots at these values.

User Jon Kartago Lamida
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