(3m^3-1/2y)(3m^3-1/2y)

Question

be as detailed as possible. thank you

Answer 1

We use FOIL method to multiply this out.

 

First terms

Outer terms

Inner terms

Last terms

 

 

Product = (3m³ * 3m³) + (3m³ * -1/2y) + (-1/2y * 3m³) + (-1/2y * -1/2y)

 

Product = 9m⁶ - (3/2)m³y - (3/2)m³y + (1/4)y²

 

Product = 9m⁶ - 3m³y + (1/4)y²

Answer 2

[3m³ - (1/2)y][3m³ - (1/2)y]

 

There are a couple of ways to look at this.

You actually have (a-b)(a-b) since the values

in each parenthesis are the same.  For that

you can use the formula

(a-b)(a-b) = a²-2ab+b²

In this case  a = 3m³ ,  b = (1/2)y

 

a²-2ab+b²

= (3m³)² -2(3m³)(1/2 y) + (1/2 y)²

= (3m³)(3m³) - 2(3)(1/2)m³y + (1/2 y)(1/2 y)

= 9m³⁺³ - 3m³y + 1/4 y²

= 9m⁶ - 3m³y + (1/4)y²

 

******************************************

 

Above is the formula method which is used in the

special case.  In general for multiplying polynomials

such as (a+b)(c+d) = ac + ad + bc + bd

In other words... multiply the first term in the 1st

parenthesis through each term in the 2nd parenthesis,

and then multiply the 2nd term in the 1st parenthesis

through each term in the 2nd parenthesis.

 

(3m³- 1/2 y)(3m³ - 1/2 y)

= (3m³)(3m³) + (3m³)(-1/2 y) + (-1/2 y)(3m³) + (-1/2 y)(-1/2 y)

= 9m⁶ - (3/2)m³y - (3/2)m³y + (1/4)y²

= 9m⁶ - (6/2)m³y + (1/4)y²

= 9m⁶ - 3m³y + (1/4)y²

 

This method works for any set of polynomials, including

the ones that fit the special patterns.