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That proof is incomplete. The problem needs f(x) to be defined for all x, but in the proof the function is not defined for x=0,1.

Yea, the "I'am not a robot" tag has an extra apostrophe or extra "a". My Grammarly senses it. 

StackExchange...huh...I answered some questions there about computer science. I didn't know they also had a math webpage.  

Please limit your questions to "math-related" questions. If you have a question that is not "math-related", you can tag your question as "off-topic" to allow other math questions to take priority. Please do not advertise products on this website...we are specialized in answering math questions only. 

If you have nowhere to post this "fat-burning" advertisement, please post your message somewhere else on social media. 

With best regards,

- PartialMathematician (Not PartialFatBurner)

We could also substitute \(y^2\) for \(x\), so we have \(16-25y^2\). Now, this looks looks like a difference of squares. We can factor it into \((4+5y)\cdot(4-5y)\). Substituting back \(x\) for \(y^2\), we have \(\boxed{(4+5\sqrt{x})\cdot(4-5\sqrt{x})}\). 

Hope this helps,

- PartialMathematician

#3 could be factored into \(4^2 - 5^2(x)\)

Wooohoooo!  Nice, Chris !

Please post this 'it-shay' somewhere else....

    DId you get hacked?   Please delete your account...and start over with a new password and ID

Solution:

\(\small \text {Powers of two (2) are the only integer numbers }\\ \hspace {1.3cm}\small \text {that cannot be expressed as a sum of two or more consecutive integers. }\\ \text { }\\ \text {To solve, find the highest power of two (2) less than } 2018. \\ \lfloor \log_2 (2018)\rfloor = \lfloor {10.9787…}\rfloor\\ 2^{10} = 1024\\ \text { }\\ \mathbf {1024} \small \text { is the largest integer less than } 2018\\ \hspace {0.9cm} \small \text { that cannot be written as a sum of two or more consecutive integers. }\\ \)

Thanks to Sir CPhill for catching my _^fuckup

GA

An altitude drawn to the hypotenuse of an isosceles right triangle always bisects the hypotenuse

So   BD = CD

And  AD forms a geometric mean of  BD and CD

So....we have this relationship

BD / AD   = AD / CD         but BD = CD  ...so...

BD / 5 = 5 / BD

BD^2   = 25

BD    =  5

But BD is 1/2 of base BC

And AD is an altitude of ABC drawn to this base

So....the area of ABC   = (1/2)BC * AD   =  5 * 5    =   25 units^2

Also

4) 4(2-a)^2 -81     is a difference of squares

[ 2 (2-a) + 9 ]  [ 2(2-a) - 9 ]   =

[4 - 2a + 9 ]  [ 4 - 2a - 9 ]    =

[13 - 2a ] [ -5 - 2a ] =       { factor a "-1" from the last term....move it to the front }

- [13 - 2a ] [ 5 + 2a ]

[ 2a - 13 ] [ 2a + 5 ]

I assume you mean the total area of the square + the triangle should be maximized

the square's area = l x l = l^2

the triangle's area = sqrt3/4 * a^2   

And    a = (5m - 4l)/3

Have to agree with above....if you want to maximize area, you would not put ANY wire in to a triangle...it would all be in the square......UNLESS you have a size for the triangle...or some sort of ratio of square to triangle....there really is no answer to this question. 

We can actually factor 3 as a difference of squares  if we note that

x  = √x * √x

16 - 25 x   =

(4 + 5√x ) ( 4 - 5√x )

We can use the Law  of Sines to solve this

Since B is the right angle   and the triangle is isosceles....then  angles  A and C = 45°

The incenter of the triangle is the point where the three angle bisectors meet

One of the triangles formed by their intersection is ΔBAL

Since B is bisected....then angle ABL    =  45°

And since A is bisected....angle BAL   = 22.5°

And angle ALB   =  180 - 45 - 22.5   = 112.5°

So

BL                       AB

_______   =      _______

sin BAL            sin ALB

BL                    4sqrt(2)

____    =        _________

sin 22.5           sin 112.5

                     4sqrt (2) sin 22.5

BL =            _______________     ≈    2.34

                          sin 112.5

Simply serach it on the internet.  LOL

a=listfor(n, 1, 23, GCD(( 9*n-2), (7*n+3))=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 41)=n=23 and the GCD of [164, 205] =41

If you use the quadratic formula you will find   a= 52/8   or -20/8

    then you would have      (a-52/8)(a+20/8)        but you will see that   a x a = a^2   NOT 4 a^2.....

        so lets multiply both terms by 2

                  (2a-52/8 * 2)( 2a+20/8 *2)

                   (2a-13)(2a+5)    

1) Factor out (a      )(3a      )

See the multiples of -8 and have them add up to -10 when one of them is multiplied by 3.

You get \((a-4)(3a+2)\)

2) The multiples of -3 are 1, -3 and -1, 3 and after trying out we get \((15x^2+3)(2x-1)\)

3) We see that we can do a difference of squares. \((4+5\sqrt x)(4-5\sqrt x)\) .

4) This equals \(4(a^2-4a+4)-81=4a^2-16a+16-81=4a^2-16a-65\)

Factor \(4a^2-16a-65=(2a+5)(2a-13)\) .

You are very welcome!

:P

(also thanks PartialMathematician for pointing out a mistake!)

4)

 Factor the following:
4 (2 - a)^2 - 81

4 (2 - a)^2 - 81 = (2 (2 - a))^2 - 9^2:
(2 (2 - a))^2 - 9^2

Factor the difference of two squares. (2 (2 - a))^2 - 9^2 = (2 (2 - a) - 9) (2 (2 - a) + 9):
(2 (2 - a) - 9) (2 (2 - a) + 9)

2 (2 - a) = 4 - 2 a:
(2 (2 - a) - 9) (4 - 2 a + 9)

Grouping like terms, -2 a + 4 + 9 = (4 + 9) - 2 a:
(4 + 9) - 2 a (2 (2 - a) - 9)

4 + 9 = 13:
(2 (2 - a) - 9) (13 - 2 a)

2 (2 - a) = 4 - 2 a:
(4 - 2 a - 9) (13 - 2 a)

Grouping like terms, -2 a - 9 + 4 = -2 a + (4 - 9):
-2 a + (4 - 9) (13 - 2 a)

4 - 9 = -5:
(-5 - 2 a) (13 - 2 a)

Factor -1 out of -2 a - 5:

-(2a + 5) (13 - 2a)

...because I misread the Q.    ~ EP

As long as we are ' fixing things'      'conjunction' ....or'contraction'?   

Example

1   (3a +2)(a-4)            the  numbers   2   and -4   must multiply to equal  '-8'  

2    (15x+3)( 2x-1)        again.....the   3  and  -1    multiply to -3   and the 15 and 2 multiply to 30

3     This one is factored as far as it can be

4    4 (a^2 -4a +4) - 81

       4a^2-16a+16 - 81

       4 a^2-16a-65          

why downvote 

You made a quick mistake forgetting the 1, but I caught it. Thank you for leading me in the right direction. :)