Factoring math problems..?

Question

solve by factoring and show all work please:

1) x(squared) = 3x

2( x(squared) - 2x = 8

Answer 1

x^2 - 3x = 0

x(x-3)=0

x = 0 or 3

x^2 - 2x - 8 = 0

(x-4)(x+2) = 0

x = 4 or -2

Answer 2

1) x(squared)=3x

you would set it up equal to zero: 0=3x-x (squared)

then you would factor out and x 0=3-x

and that would be 3

so x=3

2) x (squared) -2x=8

you would set this up equal to zero: x (squared) -2x -8 =0

then you would factor that to: (x-4)(x+2)=0

so one of those would equal zero (x-4)=0 or (x+2)=0

so the answer is 4 or -2

now you have to check them with the original equation

4 squared - 2x4 = 8

16-8=8

so 4 works

now you would check is -2 works

-2 squared -2x-2 =8

4+ 4=8

that works so your answer is

the solution set of (4, -2)

Answer 3

x² = 3x

x² - 3x = 0

x(x-3) = 0

x = 0, 3

x² - 2x = 8

x² - 2x - 8 = 0

(x - 4)(x + 2) = 0

x = 4, -2

Answer 4

the symbol "^" means to the power of.  

x^2 = 3x

Subtract 3x from both sides

x^2 - 3x = 0

factor out an "x", because this is what both terms have in common

x(x-3) = 0

the left side of the eqn is equal to zero when:

x = 0

and

x - 3 = 0

x =3

Therefore, the two solutions are when x = 0, and when x = 3.  

For the second problem:

x^2 - 2x = 8

subtract 8 from both sides:

x^2 - 2x - 8 = 0

Look at the factors of "-8", which are -1,1,-2, 2,-4,4,-8,8.  One factor must be positive, the other must be negative to get the result of "(-8)" and they must add to equal -2 (the number in front of the second term).  

The only factors that multiply to give "-8" and add to give "-2" is the factors -4, and 2.  

Therefore,

x^2 - 2x - 8 = 0 = (x-4)(x+2)

(x-4)(x+2) = 0

the left side is equal to the right side when:

x - 4 = 0 or when x = 4

and

x + 2 = 0 or when x = -2

Therefore, the two solutions are x = 4, and x = -2