# Thread: Factorise (m+1)^2 - 9

1. Originally Posted by FailInMaths
I have only learn that: ( For Factorise )

1) (a)square + 2ab + (b)square = (a+b)square
2) (a)square - 2ab + (b)square = (a-b)square
3) (a)square - (b)square = (a+b)(a-b)
Ok, but this problem is one where those cannot be applied when you subtract the 9. You have different factors.

2. Originally Posted by Unknown008
If a is m, and b is 4...

$m^2 +2m-8$

How can you replace m by a and 4 by b here?

Like this?

$a^2 +2a-2b$

This doesn't bring to (a-b)(a+b)

Like this?

$a^2 +\dfrac{ba}{2}-\dfrac{b^2}{2}$

That neither.

In short, no. If you ever got $m^2 - 9$ for example, then it would be $m^2 - 9 = (m+3)(m-3)$. But that is not the case.
Hi, can you show me the steps you do to obtain the answer completely? Maybe from there i can understand how things work, cause i just can't seem to comprehend what you guys said, sorry man, i am not advance, seek your understanding

3. The full solution:

$(m+1)^2 - 9 = m^2 + 2m + 1 - 9$

$= m^2 + 2m - 8$

$= (m+4)(m-2)$

4. Originally Posted by Unknown008
The full solution:

$(m+1)^2 - 9 = m^2 + 2m + 1 - 9$

$= m^2 + 2m - 8$

$= (m+4)(m-2)$
Got it, Thanks so much

5. This is very simple.
Just expand it out.
you should get m2+2m+1-9 = m2+2m-8
After that factorise it.
Answer should be (m+4)(m-2)

6. Originally Posted by FailInMaths
Factorise
(m+1)Square -9

My answer : (m-8)(m+8)

And also, is it true that we can't write capital letters for algebra? only small letters.

Thanks so much for helping!
Or you could use differnce of 2 squares.
Because (m+1)Square - (3) Square.
Put in the forumala (a+b)(a-b)

7. Originally Posted by Calye
Or you could use differnce of 2 squares.
Because (m+1)Square - (3) Square.
Put in the forumala (a+b)(a-b)
Yeah, i got it already, in another post, Thanks so much =)

8. No problem.

9. Originally Posted by FailInMaths
Factorise
(m+1)Square -9

My answer : (m-8)(m+8)

And also, is it true that we can't write capital letters for algebra? only small letters.

Thanks so much for helping!
I realise you think you know how to do this particular question now. However, recognising a difference of two squares should have been the prefered method of solution right from the get-go in my opinion.

$A^2 - B^2 = (A - B)(A + B)$ and in your case A = m + 1 and B = 3.

Expanding and factorising takes a lot more time, requires more skill and introduces far greater opportunity for error.

10. Convince yourself that a^2 - 9 = (a + 3)(a - 3)

(m + 1)^2 - 9 = ?
Let a = m + 1
a^2 - 9
= (a + 3)(a - 3)
Now substitute back in:
(m + 1 + 3)(m + 1 - 3)
= (m + 4)(m - 2)

11. Right. Here $(m+1)^2$ is 'a' and 9 is 'b'. So put that in identity $a^2-b^2=(a+b)(a-b)$. what you will get is:
$(m+1)^2-(3)^2$
$=[(m+1)+3][(m+1)-3]$
$=(m+4)(m-2)$
this is your solution.

12. Originally Posted by amey
Right. Here $(m+1)^2$ is 'a' and 9 is 'b'. So put that in identity $a^2-b^2=(a+b)(a-b)$. what you will get is:
$(m+1)^2-(3)^2$
$=[(m+1)+3][(m+1)-3]$
$=(m+4)(m-2)$
this is your solution.
So many replies saying exactly the same thing as all the others. Thread closed.

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