How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.
Any help would be greatly appreciated!
Thanks in advance!
What you SHOULD be doing is showing us your work, so that we can show you where you're going wrong.
Having said that...
7) $\displaystyle 3x^4 - 3x^3 - 1x^2 = 0$
$\displaystyle x^2(3x^2 - 3x - 1) = 0$
So $\displaystyle x^2 = 0$ or $\displaystyle 3x^2 - 3x - 1 = 0$
Therefore
$\displaystyle x = 0$ or $\displaystyle x = \frac{3 \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)}$
$\displaystyle x = 0$ or $\displaystyle x = \frac{3 \pm \sqrt{13}}{6}$.
So $\displaystyle A = \frac{3 - \sqrt{13}}{6}, B = 0, C = \frac{3 + \sqrt{13}}{6}$.
8) $\displaystyle 5 - x < -1$
$\displaystyle 5 + 1 < x$
$\displaystyle x > 6$.
Using interval notation, this is written as $\displaystyle x \in (6, \infty)$.
9) $\displaystyle x^3 - 16x \leq 0$
The easiest way to answer this would be to analyse the graph of $\displaystyle y = x^3 - 16x$. See where the x-intercepts are and where the graph is negative.
Well, what do you KNOW about functions? Can I presume that you know how to factor?
Function (1), $\displaystyle f(x)= -x^2(x^2- 4)= -x^2(x-2)(x+2)$ obviously has zeros at x= 0, 2, and -2 and only at those points. Only graphs A and D have that property. For x> 2, all factors, x, x-2, and x+2, are positive so, because of the "-" in front, the f(x)< 0 for x> 2 and only D has that property.
Function (2), $\displaystyle f(x)= -x^5+ 5x^3- 4x= -x(x^4- 5x^2+ 4)$$\displaystyle = -x(x^2- 4)(x^2- 1)= -x(x-2)(x+2)(x-1)(x+1)$ has zeros at x= 0, x= 2, x= -2, x= 1, and x= -1. C is the only graph that has that property.
Function (3), $\displaystyle f(x)= -x^3+ 2x^2= -x^2(x- 2)$ has zeros only at x= 0 and x= 2 (and a double root at x= 0 so the graph is tangent to the x-axis at x= 0). B is the only graph having that property.
Function (4), $\displaystyle f(x)= \frac{x^6}{2}- 2x^4=\frac{1}{2}x^4(x^2- 4)= \frac{1}{2}x^4(x-2)(x+2)$ has zeros at x= 0, x= 2, and x= -2 and, as with (1), A and D have that property. But now, for x> 2, f(x)> 0 and only A has that property.