i need help with these 4 math problems?

Question

factoring

1st: (t - 2)3 - (t - 2) = 0

2nd: (x2 - 3)2 + (x2 - 3) - 2 = 0

inequalities

3rd:x2 ≤ 9

4th:3s2 < 48

Answer 1

1st: (t - 2)3 - (t - 2) = 0

(t-2)[(t-2)^2-1]=0
(t-2)(t^2-4t+3)=0
(t-2)(t-3)(t-1)=0
t=1,2 or 3

2nd: (x2 - 3)2 + (x2 - 3) - 2 = 0
put x^2-3=t
t^2+t-2=0
(t+2)(t-1)=0
reverting
(x^2-3+2)(x^2-3-1)=0
(x^2-1)(x^2-4)=0
(x+1)(x-1)(x+2)(x-2)=0
x=+/-2 or +/-1

inequalities

3rd:x2 ≤ 9
x^2<=9
x<=3 x>=-3

4th:3s2 < 48
3s^2<48
dividing by 3
s^2<16
s<4 or s>-4

Answer 2

1)

(t - 2)^3 - (t - 2) = 0

Your first step would be to factor (t - 2) out of the terms of the left hand side.

[t - 2] [ (t - 2)^2 - 1] = 0

Now, notice the second set of brackets can be factored as a difference of squares.

(t - 2) [ (t - 2) - 1 ] [ (t - 2) + 1 ] = 0

Which we can reduce to

(t - 2) (t - 3) (t - 1) = 0

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

This can actually be treated as a quadratic, and the fact that there are multiple terms in this "quadratic" is just meant to try and throw you off. If you don't believe me, let's let u = x^2 - 3. Then

u^2 + u - 2 = 0

And let's factor this as u.

(u + 2) (u - 1) = 0

Let's replace u back. I think I made my point.

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

Look! Both of these are difference of squares, so we can factor them like so.

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

Inequalities
3) x^2 <= 9 (I use "<=" to mean "less than or equal to")

Move the 9 to the left hand side,

x^2 - 9 <= 0

And then factor as a difference of squares.

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

Thus, our critical values are -3 and 3. At these points is where the equation is EQUAL to 0 (so they will be included in our interval), but we want to determine when it is also less than 0.

What we have to do is test the areas AROUND -3 and 3. This means these intervals: (-infinity, -3], [-3, 3], [3, infinity). The great thing about this that we can test ANY number in the interval, and it will be true for that interval. What we want to test for is whether (x - 3)(x + 3) is negative.

Test x = -10000000. Then (x - 3) (x + 3) becomes (negative) times (negative) = positive. We don't want this, so we discard the interval (-infinity, -3).

Test x = 0 (since, after all, it lies between -3 and 3). Then for
(x - 3) (x + 3), we get (negative) times (positive) = negative. Which is exactly what we want! So we include the interval [-3,3].

Test x = 1000000. You'll find that it's positive, so we discard this interval.

Therefore, the solution set would be the set of all x such that x is in the interval [-3, 3].

4) 3s^2 < 48

Notice that this is strictly less than, as opposed to the other answer which was less than _or equal to_.

Divide both sides by 3, to get

s^2 < 16

Bring the 16 over.

s^2 - 16 < 0

Factor.

(s - 4)(s + 4) < 0

And now, we test points AROUND our critical values, s = -4 and 4.

Test -1000000. This gives us negative times negative = positive. Discard (-infinity, -4).

Test 0. negative times positive. keep (-4, 4)

Test 1000000. This gives us positive, so discard.

Therefore, the solution set of s is (-4,4).
These are ROUND brackets this time because they don't include -4 and 4, and they use "<" instead of "<="

Answer 3

1st t=2
2nd a foolishness
3rd x<=4.5
4th 3s2 meanswhat

Answer 4

1. 3t-6-t+2=2t-4=>t=2
2.4x-6+2x-3-2=0=>6x-11=0=>x=11/6

Answer 5

DO YOUR OWN HOMEWORK!! I ALREADY TOLD U 8 ANSWERS!!