What are the following math problems called and how do I do them?

Question

I dont want the answers, I want to learn how I go about solving them.

Math Problem #1. (x+6)(2x-8)

Math Problem #2. (x-3)(4x²-3x-6)

Math Problem #3. y²-25=0

Thank you!

Answer 1

I don't think they have "names"

Number one and two is just SIMPLIFYING.

You will either do the FOIL method or BOX method. Kind of complicated to explain online. Here's a good one: http://answers.yahoo.com/question/index;...

Number three is solving for y.

You ADD 25 to both sides to get

y^2=25

FIND THE SQUARE ROOT of both sides to get y=5

Answer 2

The WORD you are looking for is "puzzled."

Answer 3

the first one is a quadratic trinomial

expand it using FOIL (first, inner, outer, last)

the other ones....well dont ask me!!!! lol =P

Answer 4

1 and 2 are expansions - you need to multiply out the equation.

3 is solving the equation for y, possibly through a technique called completing the square.

Answer 5

its very easy actually. ok look

(x+6)(2x+8) you need to FOIL

(x+6)*2x;(x+6)*8

2x^2+12x+8x+24

2x^2+20x+24

you can then go ahead and factor (if you have to find zeros)

The first one is not factorable so you use the quad. formula

(I find discrimant first)

d=b-4ac

d=20^2-4(2)(24)

d=400-192

d=208

Now you need to write it as

-20±square root of 208/2*2

The second equation, apply the same rule as the 1st equation (not giving you answers)

the third equation solve for y

y^2-25=0

y^2=25

find square root of both sides and u got the answer.