Question about Compaq OA-LCM-14.1XGA+PA PARTS-PC8447-OMAHA
Without parenthesis, I assume you mean 12 a b^2 and 22 a^2 b.
To do a LCM, just find all the prime factors, select out enough to cover both entries, and multiply together. 12 is 3 * 2 * 2. 22 is 11 * 2. You need two 2's. So you need 2 * 2 * 3 * 11 = 132. And you need 2 a's and 2 b's. Answer 132 a^2 b^2.
Posted on Jun 13, 2008
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(a+b)2 = (a+b)(a+b) = ... ?
(a+b)2 = a2 + 2ab + b2
You can easily see why it works, in this diagram:
2. Subtract Times Subtract And what happens if you square a binomial with a minus inside?
(a-b)2 = (a-b)(a-b) = ... ?
(a-b)2 = a2 - 2ab + b2
3. Add Times Subtract And then there is one more special case... what if you multiply (a+b) by (a-b) ?
(a+b)(a-b) = ... ?
(a+b)(a-b) = a2 - b2
That was interesting! It ended up very simple.
And it is called the "difference of two squares" (the two squares are a2 and b2).
This illustration may help you see why it works:
a2 - b2 is equal to (a+b)(a-b) Note: it does not matter if (a-b) comes first:
(a-b)(a+b) = a2 - b2
The Three Cases Here are the three results we just got:
(a+b)2 = a2 + 2ab + b2 } (the "perfect square trinomials") (a-b)2 = a2 - 2ab + b2 (a+b)(a-b) = a2 - b2 (the "difference of squares") Remember those patterns, they will save you time and help you solve many algebra puzzles.
Using Them So far we have just used "a" and "b", but they could be anything.
We can use the (a+b)2 case where "a" is y, and "b" is 1:
(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1
We can use the (a-b)2 case where "a" is 3x, and "b" is 4:
(3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16
We know that the result will be the difference of two squares, because:
(a+b)(a-b) = a2 - b2
(4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4
Sometimes you can recognize the pattern of the answer:
Example: can you work out which binomials to multiply to get 4x2 - 9
Hmmm... is that the difference of two squares?
Yes! 4x2 is (2x)2, and 9 is (3)2, so we have:
4x2 - 9 = (2x)2 - (3)2
And that can be produced by the difference of squares formula:
(a+b)(a-b) = a2 - b2
Like this ("a" is 2x, and "b" is 3):
(2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9
So the answer is that you can multiply (2x+3) and (2x-3) to get 4x2 - 9
Jul 26, 2011 | Computers & Internet
Here, We deal with Some Special Products in Polynomials.
Certain products of Polynomials occur more often
in Algebra. They are to be considered specially.
These are to be remembered as Formulas in Algebra.
Remembering these formulas in Algebra is as important
as remembering multiplication tables in Arithmetic.
We give a list of these Formulas and Apply
them to solve a Number of problems.
We give Links to other Formulas in Algebra.
Here is the list of Formulas in
Polynomials which are very useful in Algebra.
Formulas in Polynomials :
Algebra Formula 1 in Polynomials:
Square of Sum of Two Terms:
(a + b)2 = a2 + 2ab + b2
Algebra Formula 2 in Polynomials:
Square of Difference of Two Terms:
(a - b)2 = a2 - 2ab + b2
Algebra Formula 3 in Polynomials:
Product of Sum and Difference of Two Terms:
(a + b)(a - b) = a2 - b2
Algebra Formula 4 in Polynomials:
Product giving Sum of Two Cubes:
(a + b)(a2 - ab + b2) = a3 + b3
Algebra Formula 5 in Polynomials:
Cube of Difference of Two Terms:
(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3
Algebra Formula 8 in Polynomials:
Each of the letters in fact represent a TERM.
e.g. The above Formula 1 can be stated as
(First term + Second term)2
= (First term)2 + 2(First term)(Second term) + (Second term)2
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