SOLUTION: Find each product 1.(3x)(4x^3)(-x^5) 2.5x^3(4x^5-2x) 3. (5x-2y)^2 4. (5x+1)(3x^2-3x-2)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find each product 1.(3x)(4x^3)(-x^5) 2.5x^3(4x^5-2x) 3. (5x-2y)^2 4. (5x+1)(3x^2-3x-2)      Log On


   

Question 88893: Find each product
1.(3x)(4x^3)(-x^5)
2.5x^3(4x^5-2x)
3. (5x-2y)^2
4. (5x+1)(3x^2-3x-2)
Found 2 solutions by malakumar_kos@yahoo.com, chitra:
Answer by malakumar_kos@yahoo.com(315) About Me  (Show Source):
You can put this solution on YOUR website!
Find each product
1.(3x)(4x^3)(-x^5)
2.5x^3(4x^5-2x)
3. (5x-2y)^2
4. (5x+1)(3x^2-3x-2)

1)(3x)(4x^3)(-x^5)
=(12x^3+1)(-x^5)
= 12x^4(-x^5)
= -12x^4+5
= -12x^9


2)5x^3(4x^5-2x)
= (5x^3)(4x^5)- (5x^3)(2x)
= (20x^3+5)-(10x^3+1)
= 20x^8-10x^4


3)(5x-2y)^2
= (5x)^2-2*5x*2y+(2y)^2
= 25x^2-20xy=4y^2



4)(5x+1)(3x^2-3x-2)
= 5x(3x^2-3x-2)+1(3x^2-3x-2)
= 15x^3-15x^2-10x+3x^2-3x-2
= 15x^3-12x^2-13x-2

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
Find each product:

1. %283x%29%284x%5E3%29%28-x%5E5%29

This can be written as:

%28+3x+%2A+4x%5E3+%2A+%28-x%5E5%29%29

Multiplying the numbers and the variables(to add the powers) we get:

+-12+x+%2A+x%5E3+%2A+x%5E5+

+-+12+x%5E9+

Hence, the solution.

2. 5x%5E3%284x%5E5-2x%29

This can be written as:

+20x%5E8+-+10x%5E4+ This is obtained by multiplying we get these terms..

Hence, the solution.

3. (5x-2y)^2

This is of the form %28a+-+b%29%5E2+=+a%5E2+-+2ab+%2B+b%5E2

Expanding this using the above formula, we get:

25x%5E2+%2B+4y%5E2+-+20xy

Hence, the solution.


4. (5x+1)(3x^2-3x-2)

Using distributive law and Multiplying terms, we get:

%2815x%5E3+%2B+3x%5E2+-+15x%5E2+-+3x+-+10x+-+2%29

==> %2815x%5E3+%2B+3x%5E2+-+15x%5E2+-+13x+-+2%29

Hence, the solution.


Regards
Chitra