# Factorise (m+1)^2 - 9

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• Feb 12th 2011, 04:20 AM
FailInMaths
Factorise (m+1)^2 - 9
Factorise
(m+1)Square -9

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

Thanks so much for helping!
• Feb 12th 2011, 04:31 AM
Unknown008
Unfortunately, no, that's not the answer.

Remember that:

$(a+b)^2 = a^2 + 2ab + b^2$
• Feb 12th 2011, 04:33 AM
HallsofIvy
Quote:

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

That would be m*m+ m*8-8*m- 8*8 and since "m*8" and "8*m" are the same thing m*8- 8*m= 0 and that is m^2- 8 ^2= m^2- 64.

Instead, you should have (m+ 1)^2= (m+1)(m+1)= m*m+ m*1+ 1*m+ 1*1= m^2+ m+ m+ 1= m^2+ 2m+ 1. Now subtract the 9.

Quote:

And also, is it true that we can't write capital letters for algebra? only small letters.
No, that is not true. What is true, and, I suspect, what you were told, is that small and capital letters are NOT interchangeable. You can have "A" and "a" but they mean different things because they are different symbols. Unless you have a good reason, it would be a bad idea to use both "A" and "a" in the same formula, since people would not be sure if you intended them to mean the same thing or not. But you have problem seen formulas such as "A= hw" or "A= (1/2)bh" for area.

Quote:

Thanks so much for helping!
• Feb 12th 2011, 04:34 AM
FailInMaths
Quote:

Originally Posted by Unknown008
Unfortunately, no, that's not the answer.

Remember that:

$(a+b)^2 = a^2 + 2ab + b^2$

Hi, but the "-9", i can't seems to find a solution
• Feb 12th 2011, 04:36 AM
Unknown008
Do the first part first.

$(m+1)^2$

After that, you'll subtract the 9.
• Feb 12th 2011, 04:38 AM
FailInMaths
Quote:

Originally Posted by HallsofIvy
That would be m*m+ m*8-8*m- 8*8 and since "m*8" and "8*m" are the same thing m*8- 8*m= 0 and that is m^2- 8

Instead, you should have (m+ 1)^2= (m+1)(m+1)= m*m+ m*1+ 1*m+ 1*1= m^2+ m+ m+ 1= m^2+ 2m+ 1. Now subtract the 9.

No, that is not true. What is true, and, I suspect, what you were told, is that small and capital letters are NOT interchangeable. You can have "A" and "a" but they mean different things because they are different symbols. Unless you have a good reason, it would be a bad idea to use both "A" and "a" in the same formula, since people would not be sure if you intended them to mean the same thing or not. But you have problem seen formulas such as "A= hw" or "A= (1/2)bh" for area.

Quote:

Originally Posted by Unknown008
Do the first part first.

$(m+1)^2$

After that, you'll subtract the 9.

Hi, so after i subtract 9, what do i do?
(a)square - (b)square = (a+b)(a-b)
(m)square +2m+1 minus 9 = (m)sqaure +2m-8
So using the formula above, answer is (m-8)(m+8)?

• Feb 12th 2011, 04:43 AM
Unknown008
Right up to:

$m^2 +2m-8$

Now, you have to factor this. The factors of -8 which gives this expression are +4 and -2.

So, $m^2 +2m-8 = (m+4)(m-2)$

Now if you expand this back, you get m^2 + 2m - 8 again

Simple, isn't it?(Smile)
• Feb 12th 2011, 04:45 AM
FailInMaths
Quote:

Originally Posted by Unknown008
Right up to:

$m^2 +2m-8$

Now, you have to factor this. The factors of -8 which gives this expression are +4 and -2.

So, $m^2 +2m-8 = (m+4)(m-2)$

Now if you expand this back, you get m^2 + 2m - 8 again

Simple, isn't it?(Smile)

Hi, but (m)square +2m-8 = (m+4)(m-2) doesn't match the formula of (a)square - (b)square = (a+b)(a-b)?
• Feb 12th 2011, 04:46 AM
HallsofIvy
Quote:

Originally Posted by FailInMaths
Hi, so after i subtract 9, what do i do?
(a)square - (b)square = (a+b)(a-b)
(m)square +2m+1 minus 9 = (m)sqaure +2m-8

Yes, that is correct.

Quote:

So using the formula above, answer is (m-8)(m+8)?
Since both Unknown008 and I have already told you that is NOT correct, why are you repeating it? I told you before that (m- 8)(m+ 8)= m*m+ m*8- 8*m- 8*8= m^- 8*8= m^2- 64 NOT m^2+ 2m- 8.

If a and b are any two numbers, then (m- a)(m+ b)= m*m+ m*b- a*m- ab= m^2+ (b- a)- ab. You want that to be the same as m^2+ 2m- 8. That is, you want b- a= 2 and ab= 8. You want two numbers whose difference is 2 and whose product is 8. What whole number factors does 8 have?

Quote:

• Feb 12th 2011, 04:49 AM
Unknown008
Quote:

Originally Posted by FailInMaths
Hi, but (m)square +2m-8 = (m+4)(m-2) doesn't match the formula of (a)square - (b)square = (a+b)(a-b)?

They shouldn't match that formula... why do you think they should? (Wondering)
• Feb 12th 2011, 04:51 AM
FailInMaths
Quote:

Originally Posted by HallsofIvy
If a and b are any two numbers, then (m- a)(m+ b)= m*m+ m*b- a*m- ab= m^2+ (b- a)- ab. You want that to be the same as m^2+ 2m- 8. That is, you want b- a= 2 and ab= 8. You want two numbers whose difference is 2 and whose product is 8. What whole number factors does 8 have?

Give me a moment for me to understand this please
• Feb 12th 2011, 04:55 AM
FailInMaths
Quote:

Originally Posted by Unknown008
They shouldn't match that formula... why do you think they should? (Wondering)

i mean, if a is m, b is 4 then should't it be (m+4)(m-4)?
• Feb 12th 2011, 04:57 AM
FailInMaths
"If a and b are any two numbers, then (m- a)(m+ b)= m*m+ m*b- a*m- ab= m^2+ (b- a)- ab."

I don't understand this part....
• Feb 12th 2011, 05:00 AM
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)
• Feb 12th 2011, 05:00 AM
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.
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