Question

1. given that $f(x) = 2x^4 - 7x^2 - 4$, which of the following statements are true? select all that apply. the remainder when $f(x)$ is divided by $(x - 2)$ is 0. there are 4 real roots. there are 2 real roots and 2 nonreal roots. since $f(-3) = 95$, $x + 3$ is not a factor. since $f(-3) = 95$, $x - 3$ is not a factor. the result when doing $\frac{f(x)}{x + 4}$ is $2x^3 - 8x^2 + 25x - 100 + \frac{396}{x + 4}$ the result when doing $\frac{f(x)}{x + 4}$ is $2x^2 - 15 + \frac{56}{x + 4}$

Explanation:

Step1: Check remainder for $x-2$

Use Remainder Theorem: $f(2)=2(2)^4-7(2)^2-4=32-28-4=0$

Step2: Find roots of $f(x)$

Let $u=x^2$, so $f(u)=2u^2-7u-4$. Factor: $(2u+1)(u-4)=0$. Solve: $u=-\frac{1}{2}$ or $u=4$. Substitute back: $x^2=-\frac{1}{2}$ (2 nonreal roots) and $x^2=4$ (2 real roots: $x=\pm2$)

Step3: Check $x+3$ as factor

Remainder Theorem: $f(-3)=95
eq0$, so $x+3$ is not a factor. For $x-3$, check $f(3)=2(81)-7(9)-4=162-63-4=95
eq0$, but the statement uses $f(-3)=95$ to conclude $x-3$ is not a factor, which is an invalid reasoning (we need $f(3)$ for $x-3$)

Step4: Divide $f(x)$ by $x+4$

Use polynomial long division or substitution. Let $f(x)=2x^4-7x^2-4$. Divide by $x+4$:
$$2x^4-7x^2-4=(x+4)(2x^3-8x^2+25x-100)+396$$

Answer:

• The remainder when $f(x)$ is divided by $(x - 2)$ is 0.
• There are 2 real roots and 2 nonreal roots.
• Since $f(-3) = 95$, $x + 3$ is not a factor.
• The result when doing $\frac{f(x)}{x+4}$ is $2x^3 - 8x^2 + 25x - 100 + \frac{396}{x+4}$