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since its a `like terms` thing and its only asking you for $x$ in the power of 2; and how do we get $x^2$? by multiplying two $x$'es right?

thus, you have yo multiply all the variables who have got an $x$ in the power of one in them:

therefore, the figure you are looking for is $-42x^2$

also you have got the $-5x^2\times 1$ as well as the $1\times -x^2$ -- add those to $-42x^2$ and you will have an answer.

first of all we are given that the roots $\in \mathbb{Z}^{++}$. going easy on this, by using vietas formula we know that the sum of the roots of that polynomial is $\frac{16}{a}$, and the product of them is $\frac{-(-k)}{a}$ where $a=1$ right?

so, this polynomial is such that it has got these three solutions: $(16,1) \ \wedge\ (8,2) \ \wedge\ (4,4)$ , therefore the sum of all those would give us $k \{ 17 \ \wedge\ 10 \ \wedge\ 8 \}$

adding them alltogether we get $35$ which when divided by three gives us $11. \overline{6}  $ or more correctly $11\frac{2}{3}$rds

we are given $f(\sqrt{x + 1}) = \frac{1}{x}$, but we know that $\sqrt{x + 1} \equiv 2$ which gives you $x+1=4 \ \ \  \therefore \ \ \  x=3 $

knowing that we can put that x on the other hand and for the first side we can either plug in $3$ for x, or just use $f(2)$ for the whole thing (which ill be using as its... easier.):

so you get:

$\boxed{ f(2)=\frac{1}{3}}  $

$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

$\begin{cases}4\left(10-2b-3c-4d\right)+b+2c+3d=4\\ 3\left(10-2b-3c-4d\right)+4b+c+2d=-10\\ 2\left(10-2b-3c-4d\right)+3b+4c+d=-4\end{cases}  $

$\begin{cases}40-8b-12c-16d+b+2c+3d=4 \\ 30-6b-9c-12d+4b+c+2d=-10\\ 20-4b-6c-8d+3b+4c+d \end{cases}  $

$ \begin{cases}-7b-10c-13d+40=4\\ -2b-8c-10d+30=-10\\ -b-2c-7d+20=-4\end{cases}  $

$ \begin{cases}-7b-10c-13d+40=4 \ \implies \ \ -7b-10c-13d+40-\left(-10c-13d\right)=4-\left(-10c-13d\right) \ \implies \ \ -7b+40=4+10c+13d \ \implies \ -7b=10c+13d-36 \ \implies \ b=-\frac{10c+13d-36}{7}  \\ -2b-8c-10d+30=-10 \\ -b-2c-7d+20=-4\end{cases}  $

$ \begin{cases}-2\left(-\frac{10c+13d-36}{7}\right)-8c-10d+30=-10\\ -\left(-\frac{10c+13d-36}{7}\right)-2c-7d+20=-4\end{cases}  $

there's quite a lot of steps, used wolfram and got:

 $ \begin{cases}\frac{-36c-44d-72}{7}+30=-10 \implies \frac{-36c-44d-72}{7}=-40 \implies -36c-44d-72=-280 \implies -36c-72=-280+44d \implies -36c=44d-208 \implies  c=-\frac{11d-52}{9}  \\ \frac{-4c-36d-36}{7}+20=-4\end{cases}  $ 

thus, 

  $ \begin{cases}\frac{-4\left(-\frac{11d-52}{9}\right)-36d-36}{7}+20=-4\end{cases} $

$ \Updownarrow  $

$ -\frac{4\left(10d+19\right)}{9}=-24  $

$ -4\left(10d+19\right)=-216  $

$ 10d+19=54  $

$10d=35$

$ d=\frac{7}{2}  $

since we know back, lets walk all the way back for, c, then b, then a:

$ c=-\frac{11d-52}{9} \Leftrightarrow c=-\frac{11\cdot \frac{7}{2}-52}{9} \implies c=-\frac{\frac{77}{2}-52}{9} \implies c=-\frac{-\frac{27}{2}}{9} \implies c=-\left(-\frac{27}{18}\right) \text{  or just }  \boxed{c=\frac{27}{18}} \Leftrightarrow  \boxed{c=\frac{3}{2}}  $

$ b=-\frac{10c+13d-36}{7} \Leftrightarrow b=-\frac{10\cdot \frac{3}{2}+13\cdot \frac{7}{2}-36}{7} \implies b=-\frac{15+\frac{91}{2}-36}{7} \implies b=-\frac{\frac{91}{2}-21}{7} \implies b=-\frac{\frac{91}{2}-\frac{21\cdot \:2}{2}}{7} \implies b=-\frac{\frac{49}{2}}{7} \implies \boxed{-\frac{49}{14}} \Leftrightarrow \boxed{b=-\frac{7}{2}} $

$a=10-2b-3c-4d \Leftrightarrow a=10-2\left(-\frac{7}{2}\right)-3\cdot \frac{3}{2}-4\cdot \frac{7}{2} \leftrightarrow a=2\cdot \frac{7}{2}-4\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a=-2\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a= -\frac{14}{-2}- \frac{9}{2}+10 \implies a= -7-\frac{9}{2}+10 \implies a= -\frac{9}{2}+3 \implies a= \frac{3\cdot \:2}{2}-\frac{9}{2} \implies \boxed{\frac{-3}{2}}  $

thus, we got:

$ d=\frac{7}{2} \ \ \ ; \ \ \  b=-\frac{7}{2}  \ \ \ ; \ \ \ c=\frac{3}{2}  \ \ \ ; \ \ \   a=-\frac{3}{2}  $

to complete our final condition of $a+b+c+d$:

$ a+b+c+d= \frac{7}{2}+\left(-\frac{7}{2}\right)+\frac{3}{2}+\left(-\frac{3}{2}\right)  $

$ \boxed{a+b+c+d =0}  $

maybe i have messed anything up in the way :\

what does $[ABC] = 240$ represent? 

here is a representative image i made:

since both diagonals cot eachother in the middle, they form 4 triangles; by looking at the image, we can say that:

$ \overline{AO}=\frac{6}{2} \ \therefore \ \overline{CO}=\frac{6}{2}  $

$ \overline{DO}=\frac{10}{2} \ \therefore \ \overline{BO}=\frac{10}{2}  $

now there are two little rules for triangles 

1) sum of any two sides $>$ third side.

2) difference of any two sides $<$ third side. ( commonly used only to prove whether  the triangle exists or not )

lets consider the $\triangle BOC $ where side $\overline{BC}=$third side (the third and largest side of the triangle, but the smallest side of the parallelogram)

1)  $ \frac{10}{2}+\frac{6}{2} > \text{ third side} $

$ 5+3> \text{ third side} $

$ 8>  \text{ third side}  $

changing the sides, would also change the 'direction' of the less-than-sign, so logically, it would become larger-than-sign:

$   \text{ third side} <8 $

so we have  $ \{  \underbrace{2}_\text{ first term },\underbrace{\frac{1}{5}}_\text{ 2nd term } , \underbrace{...}_\text{ 2 other terms },  \underbrace{a_5}_\text{ 5th term } \}  $

the formula to find any term of this geosequence is

$a_n=ar^{n-1}$ where $a_n=n^{th} term$    ;   $ a=\text{first term}$   ;   $r=\text{common ratio}$

so far we have $ a_5=2r^{5-1}   $

to find the ratio we just write this simple equation $  2\cdot \:r=\frac{1}{5}  \implies  r=\frac{1}{10} $

now that we know what the common ratio is, going back and plugging it in the formula we get:

$ a_5=2\left(\frac{1}{10}\right)^{5-1}  $

$   a_5=2\cdot \frac{1^4}{10^4}  $ 

$ a_5=2\cdot \frac{1}{10^4}  $

$   a_5=\frac{2}{10^4}  $

$ \boxed{a_5=\frac{2}{10000}}  \Leftrightarrow \boxed{ a_5=\frac{1}{5000}}  $

first of all lets evaluate a bit:

$ \frac{a+3}{a^3-a} $

$ \frac{a+3}{a\left(a^2-1\right) } $

$ \frac{a+3}{a\left(a+1\right)\left(a-1\right)}  $

now, any rational expression will be undefined when the denominator is $=\: 0$ right? because dividing by 0 ( consider it like like something which does not exist ) just makes no sense. so we note that when $a\left(a+1\right)\left(a-1\right)=0$, that expression is undefined:

$a\left(a+1\right)\left(a-1\right)=0$

$ \left(a+1\right)\left(a-1\right)=\frac{0}{a}  $

$  \left(a+1\right)\left(a-1\right)=0   $       $ \implies \begin{cases} a+1=0  \\ \ \ \ \ \text{and}  \\ a-1=0   \end{cases} \implies \begin{cases} a=-1  \\ \ \ \text{and}  \\ a=1   \end{cases}   $  

thus, whenever $\boxed{a=1,\: -1}$ the fraction becomes undefined

according to this website, https://www.mathwarehouse.com/answered-questions/multiples/what-are-the-multiples-of-21-solved.php all the multiples of 21 are:

$ 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294, 315, 336, 357, 378, 399, 420, 441, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987, 1008, 1029 $

we can right away cancel all the ones which are $< \: 500$ right?

its just easy from here -- looking from the left to the right we have to find which number has no figures $< \: 500$ :

$  \color{black}{504, 525, 546,} \color{royalblue}{567}   \color{black}{,588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987, 1008, 1029   } $

so your answer is $\boxed{\color{royalblue}{567}}$

oh lord, this is hard to swallow 

oh well, lets start shall we -- pay attention to every step:

$\Huge\text{NOTE:}$  this website kind of screwed up the colouring i do not know why but check this link for a more accurate version: http://mathb.in/59340

---------------------------------------------------------------------------------------------------------------------------------------------------

$ \color{lightskyblue}{\sqrt{60x}} \sqrt{12x}\sqrt{63x}\sqrt{42x} $

$  \color{lightskyblue}{\sqrt{4(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}  $

$ \color{lightskyblue}{ \sqrt{2^2(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}   $

$   \color{lightskyblue}{2\sqrt{(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}    $

now lets do the second one: $ 2\sqrt{(15)x}   \color{tomato}{\sqrt{12x}}\sqrt{63x}\sqrt{42x} $

$ 2\sqrt{(15)x}   \color{tomato}{ \sqrt{4(3)x}}\sqrt{63x}\sqrt{42x}  $

$ 2\sqrt{(15)x}  \color{tomato}{ \sqrt{2^2(3)x}}\sqrt{63x}\sqrt{42x}    $

$   2\sqrt{(15)x}   \color{tomato}{ 2 \sqrt{(3)x}}\sqrt{63x}\sqrt{42x}   $

now lets do the third one: $  2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{63x}} \sqrt{42x}     $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{9(7)x}} \sqrt{42x}    $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{3^3(7)x}} \sqrt{42x}    $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{(3\sqrt{(7)x}) } \sqrt{42x}    $

We are left with

$ 2\sqrt{(15)x}2 \sqrt{(3)x} (3\sqrt{(7)x})  \sqrt{42x}  $ 

firstly lets multiply the first two terms:

$ \color{darkorange}{2\sqrt{(15)x}2 \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $ 

$ \color{darkorange}{4\sqrt{(15)x} \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{3x(15x)}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{45x^2}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{9(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{3^2(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x} \Leftrightarrow \color{darkorange}{4\sqrt{3^2 \cdot x^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

$\color{darkorange}{4\sqrt{(3x)^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

$\color{darkorange}{4[(3x)\sqrt{5}]} (3\sqrt{(7)x}) \sqrt{42x}  $

$  \color{darkorange}{12x\sqrt{5}} (3\sqrt{(7)x}) \sqrt{42x}  $

now lets do work the $12x\sqrt{5} (3\sqrt{(7)x})$ :

$\color{mediumseagreen}{12x\sqrt{5} (3\sqrt{(7)x})} \sqrt{42x}$  

$\color{mediumseagreen}{36x\sqrt{7x\cdot 5} } \sqrt{42x}$

$\color{mediumseagreen}{36x\sqrt{35x} } \sqrt{42x}$

finally, we multiply them all together:

$\color{navy}{36x\sqrt{35x}  \sqrt{42x}}$

$\color{navy}{36x\sqrt{42x(35x)}}$

$\color{navy}{36x\sqrt{1470x^2}}$

$\color{navy}{36x\sqrt{(7x)^2 \cdot 30 }}$

$\color{navy}{36x(7x\sqrt{30})}$

$\color{navy}{36x\cdot x(7\sqrt{30})}$

$\color{navy}{36x^2(7\sqrt{30})}$

$\color{navy}{252x^2(\sqrt{30})}$

or just 

$\color{navy}{252x^2\sqrt{30}}$

$\color{navy}{\left(6\sqrt{7}\right)x^2\sqrt{30}}$

$\color{navy}{6\sqrt{7\cdot \:30}x^2}$
 
$ \boxed{\color{navy}{ 6\sqrt{210}x^2}} $ 

can you do anything else to this? i dont think so.
 

that is just a little off.

x^2 + bx + c = 0

we know that one solution is -7, and the other one is 8

that means:

$\left. \begin{array}
 \ x=-7 \\ x=8 
\end{array} \right\}  \implies \left. \begin{array}
 \ x+7=0 \\  x-8=0 
\end{array} \right\} $    knowing that we can write the quadratic as a factorised form$ (x+7)(x-8)  $

new lets find the quadratic:

$  (x+7)(x-8) \overset{\left(m+n\right)\left(p-q\right) = mp-mq+np-nq }{=============\Rightarrow} xx+x\left(-8\right)+7x+7\left(-8\right) $

$xx-8x+7x-7\cdot \:8$

$ \boxed{ x^2-x-56 } $

‎‎‎‎‎‎‎

i feel like i have answered this before on a different website LOL. anyway.

we can set this up as an equation, by denoting the box as any variable -- i will let it be $n$

so we have $\left(7t-10\right)\left(5t+n\right) \overset{\left(t-m\right)\left(t+n\right)=t^2+nt-mt-mn }{==========\Rightarrow } 35t^2+7nt-50t-10n $

where 

$ y=\underbrace{35t^2}_\text{quadratic term}+ \underbrace{7nt-50t}_\text{linear term}  - \underbrace{10n}_\text{constant term} $

we can see that adding $n$ makes this a trinomial -- to turn it into a binomial we have to either work the constant term or the linear term by equalising them by 0.

1) constant term:  $ 10n=0 \implies n=0 $  ;  in our question it was stated that $n\neq 0 \ \  \wedge \ \  n\in \mathbb{Q}   $ so that does not statisfy our conditions

2) linear term:  $ 7nt-50t= 0 \implies  7nt=50t \implies n=\frac{50}{7} \overset{\text{as a mixed number}}{=======\Rightarrow} 7\frac{1}{7}  $ 

that does satisfy our conditions, so that is the value of $n$.

$ n=\boxed{7\frac{1}{7}}$

$ \log _2\left(24\right)-\log _2\left(3\right)=\log _5\left(x\right)\quad \overset{\Large{\text{if} \log_a(b)=c \: \therefore \: b=a^c }}{===============\Rightarrow }\quad \:5^{\log _2\left(24\right)-\log _2\left(3\right)}=x $

$5^{\log _2\left(\frac{24}{3}\right)}=x$

$5^{\log _2\left(8\right)}=x$

$5^{\log _2\left(2^3\right)}=x$

$5^{3}=x$

$\boxed{125=x}$