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