All Answers 358150 Answers

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}$$

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

We solve:

$$\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

This will help you remember it!

https://www.youtube.com/watch?v=O8ezDEk3qCg

Have a look at this thread.

http://web2.0calc.com/questions/indices-especially-negative-indices

Thanks zegroes, bioschip shot through!  

You may want to - that's okay

BUT 

If you do eat all the icecream you will just get an enormous stomach ache!  It will not be worth it!

Anonymous has got the right idea too

10%3 = 1 ==>  10mod3 is 1 because 10 divided by 3 has a REMAINDER OF 1

11%3 = 2 ==> 11MOD3 is 2

12%3 = 0 ==> 12 mod 3 is 0

modulo

don't know the exact english word, but it's the rest of the division.

For example

10%3 = 1

11%3 = 2

12%3 = 0

That would be terrible zegroes.

A place with no maths - DEFFINITELY NOT GOOD!

Ok My turn

the roots have to be factors of 6 so I'd look at 1 first

f(1)=1+4+1-6=0 therefore (x-1) is a factor    [I am using factor theorem but it is just logical really]

I could now do algebraic division or synthetic division to find its cofactor but I'll just try easy possibilities first.

f(-2)=-8+16+-2-6=0 therefore (x+2) is another factor   

(x-1)(x+2)=x²+x-2

$$(x^3+4x^2+x-6) \div (x^2+x-2)
=x+3$$

I did the division but I don't know how to set it out in latex.  SO

$$(x^3+4x^2+x-6)=(x-1)(x+2)(x+3)$$

And that will work every time so long as the roots are integers. 

I don't know -we don't have tornados here.  

Doesn't sound good though zegroes.  

$$\sqrt{(4+16)}=\sqrt{20}=\sqrt4\times\sqrt5=2\sqrt5$$

sunday 0.5

Mon 0.5*1.5

Tues 0.5*1.5*1.5

Day 5 (friday): 0.5*1.5^5=

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{1.5}}}^{{\mathtt{5}}} = {\frac{{\mathtt{243}}}{{\mathtt{64}}}} = {\mathtt{3.796\: \!875}}$$

which is 3.8 miles

Yes the search function is problematic but the site in general is not too bad - you just have to get used to it.

I mean my last posts. The most recent in search user posts, (the olny way) does not have a title therefore cannot be got to. If that post happened some time ago it is hard to find. This is not a good design. Much prefer the other. .. and the posts are more difficult to read.

Well, I should really add a warning here.  The method is not always very useful.  The danger is that, in general, by the time you do the various substitutions to isolate a, b or c, all you get is the original equation with x replaced by a, say!!

In this case it was fairly easy to guess that 1, 2 and 3 were involved, because their product had to be 6 (the other equations help determine the signs), and I started by assuming they would be integers (I reasoned the question probably wouldn't have been posed if they weren't).  Luckily, this reasoning panned out; however, in general it won't and one could guess forever without getting anywhere!

I think we should call it :

Imperial

Minims

Mutual

Admiration

Day

Otherwise known as   "I'M  MAD"

I think we should call today.

"Mutual Admiration Day"  MAD for short.

Hi Kitty!

Fly Away Rookie Troll  !!!!

$$cosec^26a-cot^26a\\\\
=\frac{1}{sin^26a}-\frac{cos^26a}{sin^26a}\\\\
=\frac{1-cos^26a}{sin^26a}\\\\
=\frac{sin^26a}{sin^26a}\\\\
=1$$

I think they might mean this:

csc^2 (6a) - cot^2 (6a)

If so, using a trig identity, the answer is (wait for it!!)

1 = ONE = "WUN" = "UNO"

Best "math" answer of the day:

alan's answer to this.....I didn't know this could be done......(at least, I've never seen it before!!)