Questions: Find the real zeros of f. Use the real zeros to factor f. f(x)=4x^4+11x^2-3 The real zero(s) of f is/are x=-1/2, 1/2. Use the real zero(s) to factor f. f(x)=

Solution Steps

To find the real zeros of the polynomial \( f(x) = 4x^4 + 11x^2 - 3 \), we can treat it as a quadratic equation in terms of \( x^2 \). Once we find the zeros, we can use them to factor the polynomial completely.

1. Substitute \( u = x^2 \) to transform the polynomial into a quadratic equation in terms of \( u \).
2. Solve the quadratic equation \( 4u^2 + 11u - 3 = 0 \) for \( u \).
3. Substitute back \( u = x^2 \) and solve for \( x \) to find the real zeros.
4. Use the real zeros to factor the original polynomial \( f(x) \).

We start with the polynomial \( f(x) = 4x^4 + 11x^2 - 3 \). To simplify the problem, we substitute \( u = x^2 \), transforming the polynomial into a quadratic equation: \[ f(u) = 4u^2 + 11u - 3. \]

Next, we solve the quadratic equation \( 4u^2 + 11u - 3 = 0 \). The solutions for \( u \) are: \[ u_1 = -3, \quad u_2 = \frac{1}{4}. \]

We substitute back \( u = x^2 \) to find the values of \( x \):

1. For \( u_1 = -3 \): \[ x^2 = -3 \implies x = \pm \sqrt{3}i \quad (\text{complex solutions}). \]
2. For \( u_2 = \frac{1}{4} \): \[ x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \quad (\text{real solutions}). \]

Thus, the real zeros of \( f \) are \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).

Using the real zeros, we can factor the polynomial \( f(x) \): \[ f(x) = 4(x - \frac{1}{2})(x + \frac{1}{2})(x^2 + 3). \] This can be simplified to: \[ f(x) = 4\left(x^2 - \frac{1}{4}\right)(x^2 + 3) = (4x^2 - 1)(x^2 + 3). \]

Final Answer