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NEVER MIND I GOT IT 

36x^2   +  48x    + 16=

4(3x+2)(3x+2)

using the synthetic div method 

-1 ]  1  -6  -1  10   6 

       1  -1   7   -6  -4

           -7   6    4   2

ANSWER:  2 

Thanks! Also what is the correct answer to the problem?

36x^2   +  48x    + 16

\(\text{For a quadratic equation }a x^2 + b x + c = 0 \\ \\ \text{a non-negative discriminant, i.e. }\\ D=b^2 - 4 a c \\ \text{indicates real roots}\)

\(\text{the discriminant in this case is } \\ D=64+16 = 80 > 0 \\ \text{and thus we have real roots}\)

I was one off. hnn

b²*(b + 4) is      b²*b + b²*4           or             b³ + 4b²

Another example:

umm im not sure which one of these is right agghh

4b^3????

4^3?? idk >.<

Have a look at the step by step approach I've just posted to your previous question, and try the same method here.  (Hint you would start by multiplying b+4 by b^2 ).

That works too problem is she doesn't give me enough room to write it all down so i guess ill have to write really tiny lol

Here is my rough sketch:

Or, step by step like this:

Thanks I guess that's how she wants it >.< nice handwriting btw

Circumference of a circle is  pi * d        d= diameter ( 2 times the radius)

pi * (2 * 2ft)

pi * 4 = 12.57 ft

Upright:

What is your question?      Maybe.....

Perhaps this?  (sorry...sideways)

What is the area of a rectangle???

Area = Length x Width

192   =    24      x Width

Width = 192 / 24 = 8 inches.

That's not how my teacher wants me to solve it but thanks

This is long division btw (dividing polynomials)

Expand the following:
(x^5 - 8 x^4 + 22 x^3 - 27 x^2 + 20 x - 41)/(x - 4)

Express (x^5 - 8 x^4 + 22 x^3 - 27 x^2 + 20 x - 41)/(x - 4) as a sum of fractions.
(x^5 - 8 x^4 + 22 x^3 - 27 x^2 + 20 x - 41)/(x - 4) = (-41)/(x - 4) + (20 x)/(x - 4) - (27 x^2)/(x - 4) + (22 x^3)/(x - 4) - (8 x^4)/(x - 4) + x^5/(x - 4):

(-41)/(x - 4) + (20 x)/(x - 4) - (27 x^2)/(x - 4) + (22 x^3)/(x - 4) - (8 x^4)/(x - 4) + x^5/(x - 4)
= x^4 - 4x^3 + 6x^2 - 3x - 9/(x - 4) + 8

This large number(24 digits) is a prime number and is called, as you know, a "Truncatable Prime":

357,686,312,646,216,567,629,137

It has a very special and unique property which is this: If you start from the left and remove the first 3, the remaining number is STILL a prime number! If you remove the next number 5, the remaining number is still a prime number! If you continue and remove the next number 7, the remaining number is still a prime number. And if you continue with this process, the remaining number continues to be a prime number, all the way to the last digit of 7!!!. 

Isn't that remarkable?? Your sequence of primes is something like that, but much less impressive!!!.

Here's what i found about this subject: