Questions: Find the zeros of the polynomial function and state the multiplicity of each. f(x)=x^4-28 x^2+75 The smallest zero is -5 with multiplicity 1. The smaller middle zero is 3 with multiplicity 1.

Solution Steps

To find the zeros of the polynomial function \( f(x) = x^4 - 28x^2 + 75 \), we can treat it as a quadratic in terms of \( y = x^2 \). This transforms the polynomial into \( y^2 - 28y + 75 = 0 \). We can solve this quadratic equation for \( y \) using the quadratic formula. Once we find the values of \( y \), we substitute back to find the corresponding \( x \) values, which are the zeros of the original polynomial. Finally, we determine the multiplicity of each zero.

The given polynomial is a quartic polynomial: \[ f(x) = x^4 - 28x^2 + 75 \]

To find the zeros, we can substitute \( y = x^2 \). This transforms the quartic equation into a quadratic equation: \[ y^2 - 28y + 75 = 0 \]

We will solve the quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -28 \), and \( c = 75 \).

Substituting these values into the formula gives: \[ y = \frac{-(-28) \pm \sqrt{(-28)^2 - 4 \cdot 1 \cdot 75}}{2 \cdot 1} \] \[ y = \frac{28 \pm \sqrt{784 - 300}}{2} \] \[ y = \frac{28 \pm \sqrt{484}}{2} \] \[ y = \frac{28 \pm 22}{2} \]

Calculate the two possible values for \( y \):

1. \( y = \frac{28 + 22}{2} = \frac{50}{2} = 25 \)
2. \( y = \frac{28 - 22}{2} = \frac{6}{2} = 3 \)

Since \( y = x^2 \), we have:

1. \( x^2 = 25 \) gives \( x = \pm 5 \)
2. \( x^2 = 3 \) gives \( x = \pm \sqrt{3} \)

Each zero is distinct and appears only once in the factorization of the polynomial, so each has a multiplicity of 1.

Final Answer