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insens350 [35]
3 years ago
7

Which of these numbers is conposite? 4,7,23,29,41

Mathematics
1 answer:
zhenek [66]3 years ago
6 0

The composite is  4 .

I hope this helps and please mark as branilyest I need 4 more to level up in rank.


Turtle14526


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Two the two is the four.

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Factor completely.<br> 16n^6+40n^3+25
Andru [333]

Answer:Step  1  :

Equation at the end of step  1  :

 ((16 • (n6)) +  (23•5n3)) +  25

Step  2  :

Equation at the end of step  2  :

 (24n6 +  (23•5n3)) +  25

Step  3  :

Trying to factor by splitting the middle term

3.1     Factoring  16n6+40n3+25

The first term is,  16n6  its coefficient is  16 .

The middle term is,  +40n3  its coefficient is  40 .

The last term, "the constant", is  +25

Step-1 : Multiply the coefficient of the first term by the constant   16 • 25 = 400

Step-2 : Find two factors of  400  whose sum equals the coefficient of the middle term, which is   40 .

     -400    +    -1    =    -401

     -200    +    -2    =    -202

     -100    +    -4    =    -104

     -80    +    -5    =    -85

     -50    +    -8    =    -58

     -40    +    -10    =    -50

     -25    +    -16    =    -41

     -20    +    -20    =    -40

     -16    +    -25    =    -41

     -10    +    -40    =    -50

     -8    +    -50    =    -58

     -5    +    -80    =    -85

     -4    +    -100    =    -104

     -2    +    -200    =    -202

     -1    +    -400    =    -401

     1    +    400    =    401

     2    +    200    =    202

     4    +    100    =    104

     5    +    80    =    85

     8    +    50    =    58

     10    +    40    =    50

     16    +    25    =    41

     20    +    20    =    40    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  20  and  20

                    16n6 + 20n3 + 20n3 + 25

Step-4 : Add up the first 2 terms, pulling out like factors :

                   4n3 • (4n3+5)

             Add up the last 2 terms, pulling out common factors :

                   5 • (4n3+5)

Step-5 : Add up the four terms of step 4 :

                   (4n3+5)  •  (4n3+5)

            Which is the desired factorization

Trying to factor as a Sum of Cubes :

3.2      Factoring:  4n3+5

Theory : A sum of two perfect cubes,  a3 + b3 can be factored into  :

            (a+b) • (a2-ab+b2)

Proof  : (a+b) • (a2-ab+b2) =

   a3-a2b+ab2+ba2-b2a+b3 =

   a3+(a2b-ba2)+(ab2-b2a)+b3=

   a3+0+0+b3=

   a3+b3

Check :  4  is not a cube !!

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