Can you help me?

Question

1. 4x(x +1) => 3

2. Factor the following polynomial 250x^2 -2x^5 =

Answer 1

PROBLEM 1:

4x(x + 1) ≥ 3

Distribute 4x through on the left:
4x² + 4x ≥ 3

Subtract 3 from both sides:
4x² + 4x - 3 ≥ 0

Factor:
(2x - 1)(2x + 3) ≥ 0

A product will be positive if both are positive, or both are negative.

2x - 1 ≥ 0
and
2x + 3 ≥ 0

or

2x - 1 ≤ 0
and
2x + 3 ≤ 0

First set of inequalities:
2x - 1 ≥ 0
2x + 3 ≥ 0

2x ≥ 1
x ≥ 1/2
and
2x ≥ -3
x ≥ -3/2

These are both true only if x ≥ 1/2

Second set of inequalities:
2x - 1 ≤ 0
2x + 3 ≤ 0

2x ≤ 1
x ≤ 1/2
and
2x ≤ -3
x ≤ -3/2

These are both true only if x ≤ -3/2

You now have two cases where the value of x will result in a non-negative result. So the final answer is
x ≤ -3/2 or x ≥ 1/2

PROBLEM 2:

250x^2 - 2x^5

Factor out a common 2x^2
2x^2 (125 - x^3)

Now use the difference of cubes rule:
2x²(5 - x)(25 + 50x + x²)

Answer 2

1. 4x(x +1) => 3
x(x + 1) => 3/4
x^2 + x => 3/4

x^2 + x - 3/4 = 0
4x^2 + 4x - 3 = 0
x = {-4 +/- sqrt[4^2 - 4(4)(-3)]}/2(4)
x = 0.5 or -1.5

Therfore, x => 0.5

2. Factor the following polynomial 250x^2 -2x^5

250x^2 -2x^5
2x^2(125 - x^3)
2x^2(5^3 - x^3)
2x^2[(5 - x)(5^2 + 5x + x^2)]
2x^2(5-x)(25 + 5x + x^2 ANS

teddy boy

Answer 3

1.
4x(x +1) ≥ 3
=> 4x^2 + 4x + 1 ≥ 4
=> (2 x + 1)^2 ≥ 4
=> 2x + 1 ≥ 2 or 2x + 1 ≤ - 2
=> x ≥ 1/2 or x ≤ - 3/2
=> x ∈ R - (- 3/2, 1/2)

2.
250x^2 -2x^5
= 2x^2 (125 - x^3)
= 2x^2 [(5)^3 - x^3]
= 2x^2 (5 - x)(25 +5x + x^2)

Answer 4

1) 4x^2 + 4x - 3 = 0
4x^2 - 2x + 6x - 3= 0
2x(2x - 1) + 3(2x - 1) = 0
(2x + 3)(2x - 1) = 0
2x + 3 = 0, 2x - 1 = 0
x= -3/2, x= 1/2

2) this question isnt rilly clear....i think there shld be a divisor