Can someone help me out on factoring?

Question

Can you factor out #1)16x(squared)-1, #2)3x(squared)-20x-32, #3)75x(cubed)-27x {Just number the answers from 1-3}

Answer 1

1) 16x^2-1 = (4x)^2- 1^2 = (4x+1)(4x-1) using (a^2-b^2)= (a+b)(a-b)
2) 3x^2-20x -32
= 3x^4-24x + 4x-32
= 3x(x-8)+4(x-8)
= (x-8)(3x+4)

3)
75x^3 - 27x = 3x(25x^2-9) = 3x(5x+3)(5x-3)

Answer 2

1) The difference of two squares, a^2-b^2, has the factorization (a+b)(a-b). In this case, a^2 = 16x^2, and b^2 = 1, so a=4x and b=1.

2) Look for a factorization of the form
(3x+a)(x+b)
where a+3b=-20 and ab=-32. One of a and b should be positive and the other negative, since their product is negative. Try factors of 32, such as 4 and 8. You find that a=4 and b=-8.

3) Factor out a 3x, and you get 3x(25x^2 - 9). The 25x^2 - 9 part is the difference of two squares, so you apply the same technique as in problem 1.

Answer 3

#1. (16x^2-1) is a square minus a square, comes out to:
(4x-1)(4x+1)
#2. (3x^2-20x-32) equatratic equation form, comes out to:
(3x+4)(x-8)
#3. (75x^3-27x) pull out a GCF, then it is a square minus a square:
3x(25x^2-9) => 3x(5x+3)(5x-3)

Answer 4

#1. 16x^2 - 1

= (4x - 1)(4x + 1)


#2. 3x^2 - 20x - 32

= (3x + 4)(x - 8)


#3. 75x^3-27x

= 3x (25x^2 - 9)

= 3x (5x - 3)(5x + 3)

Answer 5

#1. 16x^2 - 1 = (4x+1)(4x - 1)

#2. 3x^2 - 20x - 32
= 3x^2 + 4x - 24x - 32
= 3x(x - 8) + 4(x - 8)
= (3x + 4) (x - 8)

#3. 75x^3 - 27x
= 3x(25x^2 - 9)
= 3x(5x - 3)(5x + 3)

Answer 6

1) 16x^2-1=(4x+1)(4x-1)
2) 3x^2-20-32=(3x+4)(x-8)
3) 75x^3-27x=. 3x(25x^2-9)=>3x(5x+3)(5x-3)

Answer 7

1) (4x + 1)(4x -1)
2) (3x + 4)(x - 8)
3)3x(5x + 3)(5x - 3)

Answer 8

1) (4x-1)(4x+1)
2) (3x+4)(x-8)