Question

consider the following.\\(f(x) = x^5 + x^3 - 42x\\)\\(\text{(a) find all real zeros of the polynomial function. (enter your answers as a comma-separated list. if there is no solution, enter no solution.)}\\)\\(x = -2.45,0,2.45\\)

Explanation:

Step1: Factor the polynomial

Given \( f(x) = x^5 + x^3 - 42x \), first factor out the common factor \( x \):
\( f(x)=x(x^4 + x^2 - 42) \)
Then, let \( u = x^2 \), so the quadratic in \( u \) is \( u^2+u - 42 \). Factor the quadratic:
We need two numbers that multiply to \(-42\) and add to \(1\). The numbers are \(7\) and \(-6\). So \( u^2 + u - 42=(u + 7)(u - 6) \)
Substitute back \( u = x^2 \):
\( x^4 + x^2 - 42=(x^2 + 7)(x^2 - 6) \)
So the factored form of \( f(x) \) is \( f(x)=x(x^2 + 7)(x^2 - 6) \)

Step2: Find the real zeros

To find the real zeros, set \( f(x) = 0 \):

1. For \( x = 0 \): \( 0\times(0^2 + 7)\times(0^2 - 6)=0 \), so \( x = 0 \) is a zero.
2. For \( x^2+7 = 0 \): \( x^2=-7 \), which has no real solutions since the square of a real number is non - negative.
3. For \( x^2 - 6=0 \): \( x^2 = 6 \), then \( x=\pm\sqrt{6}\approx\pm2.45 \)

Answer:

\(-\sqrt{6},0,\sqrt{6}\) (or approximately \(-2.45,0,2.45\))