Factor completely: 4x2 – 121

a. (2x – 11)(2x + 11)
b. (2x – 11)(2x – 11)
c. (4x – 121)(x + 1 )
d. (4x – 121)(x – 1)

Question

Factor completely: 4x2 – 121 

    a. (2x – 11)(2x + 11) 
    b. (2x – 11)(2x – 11) 
    c. (4x – 121)(x + 1 ) 
    d. (4x – 121)(x – 1)

diyadiya · Answer

It can be written as $$(2x)^2 - (11)^2$$

diyadiya · Answer

Use this $$(a)^2 - (b)^2 = (a+b)(a-b)$$

diyadiya · Answer

Here a = 2x & b= 11

diyadiya · Answer

So which one do you think is the answeR?

anonymous · Answer

b ?

diyadiya · Answer

Why?

diyadiya · Answer

You just have to apply it to the formula i gave you :)

anonymous · Answer

Okay...the last few problems were all perfect squares. In order to factor that you must have a + and - on both sides. Like @Diyadiya said (a+b)(a-b). Now thatyou know this you take the square root of 4 and 121. the square root of 4 is a and the square root of 121 is b. Do you understand this?

diyadiya · Answer

@heatherbenbow13 Explained well :)

anonymous · Answer

Thank you. I really hope this helps, I can't think of another way to explain it.

diyadiya · Answer

:)