Questions: Find the real zeros of the following polynomial. f(x)=x^6-3x^4-88x^2

Solution Steps

To find the real zeros of the polynomial \( f(x) = x^6 - 3x^4 - 88x^2 \), we can factor the polynomial and solve for the values of \( x \) that make the polynomial equal to zero. First, we factor out the common term \( x^2 \), and then solve the resulting quadratic equation.

The polynomial \( f(x) = x^6 - 3x^4 - 88x^2 \) can be factored by taking out the common term \( x^2 \): \[ f(x) = x^2 (x^4 - 3x^2 - 88) \]

Next, we need to solve the quadratic equation \( x^4 - 3x^2 - 88 = 0 \). We can make a substitution \( y = x^2 \), transforming the equation into: \[ y^2 - 3y - 88 = 0 \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = -88 \), we find the values of \( y \).

The solutions for \( y \) yield: \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-88)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 352}}{2} = \frac{3 \pm \sqrt{361}}{2} = \frac{3 \pm 19}{2} \] Calculating the two possible values for \( y \): \[ y_1 = \frac{22}{2} = 11, \quad y_2 = \frac{-16}{2} = -8 \] Since \( y = x^2 \), we have \( x^2 = 11 \) (which gives real solutions) and \( x^2 = -8 \) (which does not yield real solutions).

Thus, the real solutions for \( x \) are: \[ x = 0, \quad x = \pm \sqrt{11} \] Calculating \( \sqrt{11} \) gives approximately \( 3.3166 \).

Final Answer