MATH HELP!! please!!?

Question

Math help 3 question. Anyone Help??
Factor completely
1. k^4 - 5k^2 + 4
2. 6x^3 + 3x^2 - 9x =0

Solve the equation, noting any double roots.
3. x^3 + 9x = 6x^2

Thanks. I have had a lot of homework because my grandpa has been in the hospital and these are the ones i have left and i need some help. can you tell how to get the answer too so i can see how to do it. Thanks again. :)

Answer 1

1. k^4 - 5k^2 + 4

We factor this like we would a quadratic, except in this case we split k^4 into k^2 and k^2.

(k^2 - 4) (k^2 - 1)

Note that we can now factor each of those as a difference of squares. As a reminder, a^2 - b^2 factors into (a - b) (a + b).

(k - 2) (k + 2) (k - 1) (k + 1)

2. 6x^3 + 3x^2 - 9x = 0

The first thing we do is factor out the greatest common monomial. In this case it's 3x.

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

We then factor once more

3x(2x + 3) (x - 1) = 0

Solving, we have 3x = 0, 2x + 3 = 0, x - 1 = 0, which is then
x = {0, -3/2, 1}

3. x^3 + 9x = 6x^2

The first thing we have to do is move 6x^2 to the left hand side, so we get one side to be 0.

x^3 - 6x^2 + 9x = 0

Factor out the monomial first.

x(x^2 - 6x + 9) = 0

Factor again; it's a perfect square.

x(x - 3) (x - 3) = 0

Therefore, noting double roots,

x = {0, 3, 3}

Answer 2

1. k^4 - 5k^2 + 4 = (k^2-4)(k^2-1) = (k-2)(k+2) (k-1)(k+1) .
2. 6x^3 + 3x^2 - 9x =0 = 3x(2x^2 +x -3)= (2x+3)(x-1)=0
3. x^3+9x-6x^2 = x(x^2-6x+9) =x(x-3)^2
x=3 is a double root and x=0 is a simple root. .

Answer 3

1)(k^2-4)(k^2-1)
2)2x^2+x-3=0

3) x^3-6x^2+9x = 0
x^2 - 6x + 9 = 0
(x-3)^2 = 0 x=3

Answer 4

1 (k^2-4)(k^2-1)
=(k+2)(k-2)(k+1)(k-1)
k=-2,2,-1,1

2. 3x(2x^2+x-3)=3x(2x+3)(x-1)=0
x=0, 1, -3/2

3. x^3+9x-6x^2=x(x^2-6x+9)
=x(x-3)^2
x=0, 3

Answer 5

3. x^3 + 9x = 6x^2

3. x^3-6x^2+ 9x =0

3x(x^2-6x+9)=0

(x^2-6x+9)=0

x^2-3x-3x+9=0

x(x-3) -3(x-3)=0

(x-3)(x-3)=0

x=3