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!

Printable View

- Feb 12th 2011, 03:20 AMFailInMathsFactorise (m+1)^2 - 9
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! - Feb 12th 2011, 03:31 AMUnknown008
Unfortunately, no, that's not the answer.

Remember that:

$\displaystyle (a+b)^2 = a^2 + 2ab + b^2$ - Feb 12th 2011, 03:33 AMHallsofIvy
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.

**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, 03:34 AMFailInMaths
- Feb 12th 2011, 03:36 AMUnknown008
Do the first part first.

$\displaystyle (m+1)^2$

After that, you'll subtract the 9. - Feb 12th 2011, 03:38 AMFailInMaths
- Feb 12th 2011, 03:43 AMUnknown008
Right up to:

$\displaystyle m^2 +2m-8$

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

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

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

Simple, isn't it?(Smile) - Feb 12th 2011, 03:45 AMFailInMaths
- Feb 12th 2011, 03:46 AMHallsofIvy
Yes, that is correct.

Quote:

So using the formula above, answer is (m-8)(m+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:

Please enlighten me

- Feb 12th 2011, 03:49 AMUnknown008
- Feb 12th 2011, 03:51 AMFailInMaths
- Feb 12th 2011, 03:55 AMFailInMaths
- Feb 12th 2011, 03:57 AMFailInMaths
"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, 04:00 AMFailInMaths
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, 04:00 AMUnknown008
If a is m, and b is 4...

$\displaystyle m^2 +2m-8$

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

Like this?

$\displaystyle a^2 +2a-2b$

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

Like this?

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

That neither.

In short, no. If you ever got $\displaystyle m^2 - 9$ for example, then it would be $\displaystyle m^2 - 9 = (m+3)(m-3)$. But that is not the case.