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.Com
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) (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) (Show Source): You can put this solution on YOUR website!
Find each product:
1.
This can be written as:
Multiplying the numbers and the variables(to add the powers) we get:
Hence, the solution.
2.
This can be written as:
This is obtained by multiplying we get these terms..
Hence, the solution.
3. (5x-2y)^2
This is of the form
Expanding this using the above formula, we get:
Hence, the solution.
4. (5x+1)(3x^2-3x-2)
Using distributive law and Multiplying terms, we get:
==>
Hence, the solution.
Regards
Chitra
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