Multiplying polynomials?

Question

I need some help explianing to me how to do these kinds of problems.

1. multiply (3x+9)^2

2. (4m-3n)(4m+3n)

with the answers and steps to these I am hopeing that I can figrue out the rest on my own by refferring back to these. Thanks.

Answer 1

1)(a+b)^2=(a)^2 + (b)^2 + 2ab.....
(3x+9)^2=(3x)^2 + (9)^2 + 2 (3x) (9)
=9x^2 + 81 + 54x

2)(a+b)(a-b)=a^2 + b^2
(4m-3n)(4m+3n)=(4m)(4m) + (4m)(3n) - (3n)(4m) - (3n)(3n)
= (4m)^2 - (3n)^2
= 16m^2 - 9n^2

Answer 2

1. it's like (a + b)^2 = a^2 + 2ab + b^2
you'll get (3x + 9)^2 = 9x^2 +54x + 81

2. it's like the reverse of (a^2 - b^2) = (a + b)(a - b),
you'll get (4m - 3n)(4m + 3n) = 16m^2 - 9n^2

It used to be called remarkable identities in the remote past. Hope it helps.

Answer 3

Use the FOIL method. Multiply the

First terms
Outside terms
Inside terms
Last terms

(3x + 9)² = (3x + 9)(3x + 9) = (3x)(3x) + (3x)(9) + (9)(3x) + 9*9
= 9x² + 27x + 27x + 81 = 9x² + 54x + 81


(4m - 3n)(4m + 3n) = (4m)(4m) + (4m)(3n) - (3n)(4m) - (3n)(3n)
= 16m² + 12mn - 12mn - 9n² = 16m² - 9n²

Notice that the two middle terms cancelled out.

Answer 4

1. (3x+9)^2 is in the form (a+b)^2
ie, a^2+2ab+b^2
So, (3x+9)^2 = (3x)^2 + 2*3x*9 + 9^2
= 9x^2 + 54x + 81

2. (4m-3n)(4m+3n) = (4m)^2 - (3n)^2
= 16m^2 - 9n^2

Answer 5

1. (3x+9)^2 is of the form
(a+b)^2= a^2 +2ab + b^2
so we will get: 9x^2 + 2*3*9x +81
= 9x^2 + 54 + 81

2. (4m - 3n)(4m + 3n) is of the form
(a - b)(a+b) = a^2 - b^2
so we get : 16m^2 - 9n^2

Answer 6

1. 9x^2+54x +81
2. 16m^2-9n^2