multiply (ab - 2) (ab + 2) and factoring

Question

1.  (ab -2) (ab +2) = ab^2 + 4   (is this correct)

 

2.  Can 8a^3 -27 be factored?  I cannot find a way to do it

 

3.  Can x(x -2)(x +3) = 0 be solved?  I did x=2 and y = -3 and nothing fits to have the problem still equal zero.

 

4.  can this problem be simplified, if so how?   Or is it undefined, if so why?

 

           a^2^m b^m^+^1

                  a^m b

Answer 1

1. (ab -2) (ab +2) = ab^2 + 4 (is this correct)

 

No: (ab -2) (ab +2) = (ab)^2 - 2^2 = a^2b^2 - 4

2. Can 8a^3 -27 be factored? I cannot find a way to do it

 

Use Difference of Cubes: (x^3 – y^3) = (x – y)(x^2 + xy + y^2)

8a^3 - 27 = (2a)^3 - 3^3 = (2a - 3)(4a^2 + 6a + 9)

3. Can x(x -2)(x +3) = 0 be solved? I did x=2 and y = -3 and nothing fits to have the problem still equal zero.

 

Use the Zero Product Property (If the product of numbers is 0, then at least one of them has to be 0).

 

x(x -2)(x +3) = 0

==> x = 0 or x-2 = 0 or x+3 = 0

==> x = 0 or x = 2 or x = –3

4. can this problem be simplified, if so how? Or is it undefined, if so why?

(a^2^m b^m^+^1)/(a^m b)

 

"^+^1" doesn't make sense; I think it should be "^(+1)".

 

Because PEMDAS includes a left-to-right rule, "a^2^m" means "(a^2)^m".

 

((a^2)^m (b^m)^(+1))/(a^m b)

= (a^(2m) b^m)/(a^m b)

= a^(2m-m) b^(m-1)

= a^m b^(m-1)

Answer 2

1.  It  should be ab^2-4.  To see this, just multiply it out term by term.  ab*ab = (ab)^2; ab*2 = 2ab; -2*ab = -2ab; -2*2 = -4.  So (ab)^2+2ab-2ab-4 = (ab)^2-4.

 

2.  Yes, it factors into (2a-3)*(4a^2+6a+9).  For this, notice that the first two terms must multiply to equal 8 and the last two have to multiply to -27.  So we only have a few choices to try.

 

3. x=0, x=2, and x=-3 are solutions.  This is because these values make each of your 3 terms zero thus making the entire expression x*(x-2)*(x+3) zero.

 

4. You're going to have to clarify this expression.  a^(2^m) is very different from (a^2)^m.  Also, b^m^+^1 doesn't make sense.