I need to factor 4x^2-1?

Question

School just started back up and I forgot a lot over the summer. Thanks! I would appreciate it if someone could explain this and work it out. I am also having trouble with this one: Solve: 8-2y<(or equal to)6

Answer 1

4x^2 - 1 Factor using A*C method ax^2 + bx + c No coefficient for b

4x^2 + 2x - 2x - 1 A*C = 4. 2*2 Also = four but also 2-2 = 0 which is b

2x(2x + 1) - 1(2x + 1) Factor 4x^2 + 2x and 2x - 1.

(2x + 1)(2x-1) ANSWER

Note that 2x(2x + 1) - 1(2x + 1). 2x + 1 is in both (...). That should always be. (2x + 1) ( ? ) Once you take out 2x + 1, you are left with

2x(...) - 1(...) What is left over? 2x - 1. Therefore your answer is (2x + 1)(2x - 1)

Always FOIL to make sure you factored correctly.

8 - 2y ≤ 6 Subtract 8 from both sides to solve for y

8 - 8 - 2y ≤ 6 - 8

-2y ≤ -2 Divide -2 from both sides.

y ≥ 1 ANSWER *Flip sign when dividing by a negative (in inequalities)*

Answer 2

Easy (2x+1) (2x-1).  When you have (ax^2-b) and both A and B can be square rooted, the answer is (rootax-rootb)(rootax+rootb).

Answer 3

1.This is a quadratic equation.

   4x^2 -1 = 0

=4x^2 +2x - 2x  -1=0

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

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

Therefore x = either 1/2 or -1/2

2.

  8-2y&lt;(or equal to) 6

=2&lt;(or equal to)2y

=1&lt;(or equal to)y

Answer 4

(2x - 1)(2x + 1)

8 - 2y ≤ 6

2 ≤ 2y

1 ≤ y

y ≥ 1

Answer 5

4x^2-1

(2x - 1)(2x + 1)

8-2y&lt;(or equal to)6

2&lt;(or equal to)2y

y&gt;(or equal to)1