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• October 31st 2010, 02:47 AM
Roxas
1) Expand and Simplify

(4x-3)(x+5)

2) Prove that

(n+5)^2 - (n+3)^2 = 4(n+4)

3)

a) Expand and simplify (2n+1)^2

b) Prove that the square of any odd number is always 1 more than a multiple of 8.

4) Expand and simplify (x-3)(2x+1)

5)

a) Expand and simplify (x+y)(x-y)

b) Using your answer to part (a). or otherwise, find the exact value of
780^2 - 220^2
• October 31st 2010, 02:50 AM
Pim
What have you tried yourself? What do you have problems with?
• October 31st 2010, 03:01 AM
Roxas
Could you perhaps give a brief step by step guide as to what to do?

I missed alot of work on this and i'm struggling. Thanks.
• October 31st 2010, 03:06 AM
Unknown008
1) Multiply 1st term in first bracket by 1st term in second bracket
Add to this, the product of the 1st term in the 1st bracket and the 2nd term in the 2nd bracket
Add to this, the product of the 2nd term in the 1st bracket and the 1st term in the 2nd bracket
Add to this, the product of the 2nd term in the 1st bracket and the 2nd term in the 2nd bracket

2) Can you expand the two brackets on the left?

We'll do the others after those.
• October 31st 2010, 03:19 AM
Roxas
1)

4x * x = 4x^2
4x * 5 = 20x
-3 * x = -3x
-3 * 5 = -15

= 4x^2 + 20x -3x -15

Correct so far?
• October 31st 2010, 03:20 AM
Unknown008
Yes, you simplify now, taking the part containing: 20x - 3x
• October 31st 2010, 03:22 AM
Roxas
8x + 17x - 15?
• October 31st 2010, 03:23 AM
Unknown008
Great! Next number now. (Smile)
• October 31st 2010, 03:28 AM
Roxas
How do you expand the two brackets on the left? >_<
• October 31st 2010, 03:32 AM
Unknown008
Just like you did for the first one.

(n+5)^2 - (n+3)^2 = (n+5)(n+5) - (n+3)(n+3)

You'll notice that if you have something of the type:

$(a+b)^2$

It expands in the fom:

$a^2 + 2ab + b^2$

But for the time being, expand it like you did for 1)
• October 31st 2010, 03:33 AM
Roxas
Whilst doing this, do I ignore the powers?
• October 31st 2010, 03:35 AM
Unknown008
You already took the powers into consideration.

Doesn't

$5^2 = 5\times5 = 25$

$x^2 = x \times x$

Similarly,

$(n+5)^2 = (n+5)(n+5)$
• October 31st 2010, 03:35 AM
Roxas
(n+5)^2 - (n+3)^2

n x n = n^2
n x 3 = 3n
5 x n = 5n
5 x 3 = 15

n^2 + 3n + 5n + 15
= N^2 + 8n + 15

Did i do it right? :S
• October 31st 2010, 03:37 AM
Unknown008
No, you didn't get my point.

First expand (n+5)^2.

Then expand (n+3)^2.

Lastly, subtract the two expansions.
• October 31st 2010, 03:40 AM
mr fantastic
Quote:

Originally Posted by Roxas
(n+5)^2 - (n+3)^2

n x n = n^2
n x 3 = 3n
5 x n = 5n
5 x 3 = 15

n^2 + 3n + 5n + 15
= N^2 + 8n + 15

Did i do it right? :S

No. If you want to do it by expanding each term, then you need to recall that (a + b)^2 = a^2 + 2ab + b^2. After correctly expanding, you then need to correctly take the difference. It should be clear, for example, that the final answer will NOT include n^2 ....

I suggest you get a copy of the class notes that you missed and review them carefully. Your textbook will also have exmaples that you should carefully review.
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