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What is the answer to (3x+5)(3x-5)? I know it is simple but I can't seem to get it, and unfortunatly I lost my math book. Please help!

 Jan 22, 2015

Best Answer 

 #3
avatar+118677 
+8

You cannot solve it Tenacious because there is no equal sign :)

The asker want to expand and simplify it.

(3x+5)(3x-5)

you can expand it the long way but you should learn to recognise this as the difference of 2 squares.

The brakets are the same but one has a minus and the other a plus.  (Thay are a conjugate pair)

 

the answer will be      $$(3x)^2-5^2=9x^2-25$$

 Jan 23, 2015
 #1
avatar+226 
0

What is the answer to (3x+5)(3x-5)?

Firstly, are you trying to expand these brackets or solve this for x?

 Jan 22, 2015
 #2
avatar
0

9x-15 😃 ....

 Jan 22, 2015
 #3
avatar+118677 
+8
Best Answer

You cannot solve it Tenacious because there is no equal sign :)

The asker want to expand and simplify it.

(3x+5)(3x-5)

you can expand it the long way but you should learn to recognise this as the difference of 2 squares.

The brakets are the same but one has a minus and the other a plus.  (Thay are a conjugate pair)

 

the answer will be      $$(3x)^2-5^2=9x^2-25$$

Melody Jan 23, 2015
 #4
avatar+226 
+5

Yes, unsolvable for x as it is, I was hoping for more information if they replied.

(3x+5)(3x-5)   If I expand this I get;

3x2-15x+15x-25

3x2-25      How you have got to your answer, I have no idea.

I have heard the term "difference of two squares" but that's my only exposure to it.

 Jan 23, 2015
 #5
avatar+118677 
+5

It is not $$3x^2$$

it is     $$(3x)^2=3x*3x=9x^2$$

 

Difference of 2 squares            $$\boxed{a^2-b^2=(a-b)(a+b)}$$

It is difference because it is subtraction.

Here are some for you

$$\\simplify\\
1) (a+3)(a-3)\\
2) (2n-4)(2n+4)\\
3) (5-2t)(5+2t)\\\\\\
factorize\\
4)\;\; x^2-16\\
5) \;\;9-4m^2\\
6)\;\;16x^2-a^2b^2$$

 Jan 23, 2015
 #6
avatar+226 
+5

I don't really understand this at the moment and haven't got the time to get my head into it so I'll have to come back to it and the other factorising questions you set on the other post in a few days. I'll message you when I've done them rather than you having to check. Thanks for the challenges.

 Jan 23, 2015
 #7
avatar+118677 
+5

okay, not a problem Tenacious 

 Jan 24, 2015

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