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Arrange each term in each binomial in order of degree from greatest to least. The degree of a binomial is the exponent attached to the term. For example, 4x^2 is a second degree term.
Multiply each term in the binomial that is being subtracted by -1 to turn it into an addition problem. For example, the problem (8x^2 + 8) - (x^2 - 2) becomes (8x^2 + 8) + (-x^2 + 2).
Combine like terms. In the example problem, the x^2 terms are combined and the constant terms are combined, yielding (8x^2 + 8) + (-x^2 + 2) = 7x^2 + 10.
Understand the F.O.I.L. method. F.O.I.L. is an acronym standing for first, outside, inside and last. It means that you multiply the first number of the first binomial by the first number of the second, then the numbers on the outside (the first term of the first binomial by the second term of the second binomial) and so on. This ensures that both numbers in the first binomial are multiplied by both numbers in the second.
Use the F.O.I.L. method to multiply the two binomials together. For example, (3x + 4)(3x - 4) = 9x^2 +12x - 12x - 16. Notice that -12x is the product of the outside terms and -16 is the product of the last terms, 4 and -4.
Simplify. There will almost always be like terms to combine. In the example, 12x and -12x cancel out, yielding the answer 9x^2 - 16.
Use the distributive property to divide both terms in the binomial by the monomial divisor. For example, (18x^3 + 9x^2) / 3x = (18x^3 / 3x) + (9x^2 / 3x).
Understand how to divide by a term. If you are dividing a higher order term by a lower order term, you subtract the exponent. For example, y^3/y = y^2. The number part of each term is handled like any other division problem. For example, 20z / 4 = 5z.
Divide each term in the binomial by the divisor; (18x^3 / 3x) + (9x^2 / 3x) = 6x^2 + 3x.
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(a+b)2 = (a+b)(a+b) = ... ?
The result:
(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) = ... ?
The result:
(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) = ... ?
The result:
(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.
Example: (y+1)2
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
Example: (3x-4)2
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
Example: (4y+2)(4y-2)
We know that the result will be the difference of two squares, because:
(a+b)(a-b) = a2 - b2
so:
(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
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