MaxWong

Your question is answered here in 2019: 

https://web2.0calc.com/questions/please-help-thank-you_18

Let q(x) be the quotient when f(x) is divided by x^2 - 1.

Since the divisor is quadratic, the degree of the remainder must be at most 1. Let ax + b be the required remainder.

Then $f(x) = (x^2- 1)q(x) + (ax + b)$.

Note that $f(1) = (1^2 - 1)q(1) + a + b = a + b$, but on the other hand, we can find f(1) by substituting x = 1 into $f(x) = x^{10}+5x^9-8x^8+7x^7-6x^6+12x^5-4x^4-8x^3+12x^2-5x-5$ directly, and that gives f(1) = 1.

So a + b = 1. Similarly, we can get -a + b = -21 by considering f(-1).

Now can you solve for the values of a and b with $\begin{cases}a + b = 1\\-a + b = -21\end{cases}$? After that, the remainder is just ax + b, with the values of a and b you got.

$\quad g^4 + 12g^2 + 9 - 10g^2 + 8\\ = g^4 + 2g^2 + 17\\ =g^4 + 2g^2 + 1 + 16\\ $

You know that $(g^2 + 1)^2 = g^4 + 2g^2 +1$. Can you find the answer from here?

We can view this geometrically. Let x and y be the chosen numbers. We can plot (x, y) inside the unit square with vertices (0, 0) and (1, 1).

The constraints are $\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23$ and we are basically finding the probability that the point (x, y) falls into the region described by the compound inequality. We can do so by considering the area of the region enclosed by $\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23$ and the unit square.

Let $\Omega = ​​\left\{(x, y) \in [0, 1]^2 : \dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23\right\}$. Here's a plot of $\Omega$:

So the probability is just $\operatorname{Area}(\Omega) = 1 - 2\times \dfrac{\dfrac23 \times \dfrac23}2 = \dfrac59$

The normal vector of the plane is $3\vec{i} + 2\vec{j} + \vec{k}$.

Now, we calculate the vector from (0, -1, -3) to (-4, -2, -2).

$\vec v = (-4, -2, -2) - (0, -1, -3) = (-4, -1, 1) = -4\vec i - \vec j + \vec k$.

We project that onto the normal vector to get the normal component of the vector, and find the length of the normal component to find the height. Can you proceed from here?

The object is 148 feet above the ground, so let h = 148:

$148 = -16t^2 + 137t + 100\\ 16t^2 - 137t + 48 = 0$

You can now use quadratic formula to solve for the two values of t.

Note that $3y^2 - y - 24 + 10 = 3y^2 - y - 14$.

Then, note that -y = 6y - 7y, so we have:

$\begin{array}{cl} & 3y^2 - y - 14\\ =& 3y^2 + 6y - 7y - 14\\ =& 3y(y + 2) - 7(y + 2) \end{array}$

Now you can take out the common factor (y + 2) and compare with what the question gave you. Can you proceed from here?

We want 4x^2 + 9y^2, which is actually just (2x)^2 + (3y)^2, and the first equation looks kinda like it so we may try to expand the left-hand side of the first equation.

Using the identity (a - b)^2 = a^2 - 2ab + b^2,

$\begin{array}{rcl} (2x - 3y)^2 &=& 4\\ (2x)^2 - 2(2x)(3y) + (3y)^2 &=& 4\\ 4x^2 - 12xy + 9y^2 &=& 4 \end{array}$

We are almost there. The term we don't want is -12xy, but we know exactly what the value of -12xy is. Can you proceed from here?

The complexity of this problem is not the same level as a homework problem, so I will work it out in full details.

Using trinomial theorem (which is just a generalization of binomial theorem), we have 

$\left(2z - \dfrac1{\sqrt z} + z^3\right)^9 = \displaystyle \sum_{\substack{a + b + c = 9\\a,b,c\text{ nonnegative integers}}} \binom{9}{a,b,c} (2z)^a \left(-\dfrac1{\sqrt z}\right)^b (z^3)^c$

where $\displaystyle\binom{9}{a,b,c}$ denotes the trinomial coefficient $\displaystyle\binom{n}{a,b,c} = \dfrac{n!}{a! b! c!}$.

So basically, we first have to find nonnegative integers a, b with a + b <= 9 where a - b/2 + 3(9 - a - b) = 0.

If we move b/2 to the other side of the equation, we get b/2 = a + 3(9 - a - b). Since the right-hand side is an integer, the left-hand side, i.e., b/2, must also be an integer. So we can brute-force and try b = 0, 2, 4, 6, 8 and see which one works. Trying one by one gives a = 10, b = 2 or a = 3, b = 6 as the only solutions with b in the range 0 <= b <= 9. Since a + b <= 9, we can reject the first solution. Doing so, we have $\begin{cases}a = 3\\b = 6\end{cases}$ as the only nonnegative integer solution to the equation.

Now, the constant term is $\displaystyle \binom{9}{3,6,0} (2z)^3 \left(-\dfrac1{\sqrt z}\right)^6(z^3)^0 = \binom{9}3 2^3 = 672$.

This one is a similar problem: https://web2.0calc.com/questions/help-polynomials_9#r2