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What is the value of k if x3 + kx2 + 7x + 5 is divided by x + 6 and gives a remainder of −1?

I am going to use remainder theorum.  :)

Here is one of many references for it if you want to take a look.

Hello 

berea este buna 

(1/2) log₃ p   + (2) log₉ p  =   [ use change of base  to re-write this]

(1/2)[log p] / [log3]   + 2[logp] / [log 9]  =

(1/2) [log p] /[ log 3]  + 2[log p] / [ log(3 * 3)]  =

(1/2) [ log  p] /  [log 3]   +  2 [log p] / [ log3 + log3] =

( [logp] [log3 + log3] + 4[log p] [ log 3] ) / ( 2 log (3)* [2 log3)] ) =

( [log p] [ log3 + log 3 + 4log 3 ] ) / ( 2 log 3 * [ 2 log 3 ])  =

([log p] [ 6 log 3] ) / [ (2log 3) (2log 3)]  =      ......( 6log 3 / 2 log 3   =  3 )

[ log p] [3] / [2 log 3] =

(3/2) [log p / log 3]  =

(3/2) log₃ p

Hello ;))))))))) 

Hello:)

Ummmm, awkward...

But that's beer...

And if it is an option, it's a terrible one. 

2/3x + 1/3 + 2 1/2x - 1 - 3/5x - 3/5 =2

Combine like terms.

The decimals are approximate, the numbers would have repeated for ages otherwise.

2.57x - 1.26 = 2

Add 1.26 to both sides.

2.57x=3.26

2.57x=3.26 = x=(326/257)

Why  . . .  Why you hurt my feelings :(

 Oh it is such a cold, cold world we live I can’t stand it any more  . . .

I am going to top myself and it will be YOUR fault I hope you are happy about it and can live with yourselves. So there!

;))))))))))))))))))))))) 

When the polynomial P(x) = 3x^3 + kx^2 + 45x + 25 is divided by x + 5, the remainder is 0. Which of the following is also a factor of P(x)?

Chris has used the factor theorem.

I will try and walk you through why it works :)

If the remainder is 0 when P(x) is divided by x+5 that must mean that   x+5 is a factor of  P(x)

so     P(x) = (x+5) (some other polynomial)

If this is true then  P(x) must equal zero when   x+5=0     and that happens when  x=-5

so  P(x) must =0  when  x=-5

that is       P(-5)=0

Now I am up to where Chris started.  :)

It will make life easier for you if you work out what I am saying.  The more things you understand the better it is for you :)

\(\text{ (x+y=53) just put any x and y whose sum is 53 and you wil get the same answer as} 53^3\)

\((x+y)^3=(x+y)(x+y)(x+y)=x^3 + 3x^2y + 3xy^2 + y^3\)

Nicely put Rarin! 

K*****g urself is NOT an option. Also, think of all the  opportunities ur parents already have had to eliminate you if they wanted. 

One special thing for you good sir! I'd like to tell you a secret. You are the absolute worst. You sir, should go practice what you preach. 

Don’t listen to these kids, k*****g yourself is a solution. If you want to make a statement in your death then do it in a messy way so your parents will know how much you hate them. Brain splatter, via a shotgun, on the walls and ceiling is a good way. That is a noisy method.

For a quieter way, I suggest cutting your wrists and bleeding out on the carpet in the living room – do it at night so they can see your corps in the morning. They may hate you as much as you hate them so don’t be too surprised if they think cleaning up your last mess was well worth getting rid of you.

Let's use 4 as an example. The sqrt of 4 is what? 2. Why? Because 2*2=4.

1/2 log base 3 of p + 2 log base 9 of p

\(\begin{array}{rcll} p &=& p \\ 3^{\log_3{(p)}} &=& 9^{\log_9{(p)}} \qquad & | \qquad \log_3{()} \\ \log_3{ ( 3^{\log_3{(p)}} ) }&=& \log_3{ ( 9^{\log_9{(p)}} ) } \\ \log_3{(p)}\cdot \log_3{( 3 )}&=& \log_9{(p)}\cdot \log_3{( 9 )} \qquad & | \qquad \log_3{(3)} = 1\\ \end{array}\\ \boxed{~ \begin{array}{rcll} \log_3{(p)}&=& \log_9{(p)} \cdot \log_3{( 9 )} \qquad & | \qquad \log_3{( 9 )} = \log_3{( 3^2 )} = 2\cdot \log_3{(3)} = 2 \\ \log_3{(p)}&=& \log_9{(p)} \cdot 2 \\ \log_9{(p)} &=& \frac{ \log_3{(p)} }{2} \end{array} ~}\)

\(\begin{array}{rcll} \frac12 \cdot \log_3{(p)} + 2\cdot \log_9{(p)} &=& \frac12 \cdot \log_3{(p)} + 2\cdot \frac{ \log_3{(p)} }{2} \\ &=& \frac12 \cdot \log_3{(p)} + \log_3{(p)} \\ &=& \frac32 \cdot \log_3{(p)} \\ \end{array}\)

Let's assume that we can represent your speed at 1 PM   by the point (1,60)   on a graph and that your speed at 2 PM  can be represented by the point (2, 70).....

Since the time [ represented by all the x values between x =1 and x = 2 ]  is continuous between these two points, an associated speed at each time [represented by a y value] can be associated wth that particular time. And a continuous curve [ of some type] consisting of all these (x,y) values can be drawn connecting the two points mentioned above.......And since the curve is continous, it is obvious that no matter how fast you travel or how slowly you travel between x = 1 and x = 2  [ 0 being the lower bound, of course] that at some x value, your speed will = 65 mph

In strict math terms this says that, given  that  f(a)  and f(b)  exist on some continuous curve.......then  that curve will, at the very least,  take on all y values between f(a) and f(b)

BTW  - This is known in Calculus as the " Intermediate Value Theorem"

Lets see... You said only using six matchsticks. You never said I couldn't break them apart! e.e Break the matchsticks until you have 12 pieces and make 4 triangles.

I've been wondering if after all these years you'd like to meet...

Thanks CPhill and Guest !

If x + 5  is a factor, then -5 is a root.....so we can find the value  of "k"  as follows:

3(-5)^3 + k(-5)^2 + 45(-5) + 25  = 0

-375  + 25k - 225 + 25 = 0

25k = 375 + 225 - 25

25k = 575

k = 23

And using sythetic division, we have

-5   [ 3       23       45       25  ]

                -15      -40       25   

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

       3           8         5        0

So......the  remaining polynomial  is    3x^2 + 8x + 5

Setting this to 0 and factoring, we get :

(3x + 5) (x + 1)  = 0

(C)  is the answer