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C^ (2) - D^(2) Factors to (C -D )(C + D)

If we apply FOIL to these bracketed terms. (C -D )(C + D), then we have

F ; C^(2)

O = CD

I = -DC

L = -D^(2)

'Stringing out'

C^(2) + CD - DC - D^(2)

NB Remember CD= DC ; just like 2 x 3 = 6 & 3 x 2 =6

Hence

C^(2) + CD - CD _ D^(2)

Adding terms we have C^(2) - D^(2)

NB THe (+)CD - CD = 0

This is the inverse function, done to show how C^(2) - D^(2) factors.

NB Remember two squared terms with a negative(-) between WILL Factor.

However, two squared with a positive(+) between them does NOT factor.

As a n example, take the Pythagorean Eq'n.

h^(2) = x^(2) + y^(2)

This does NOT factor .

However,

h^(2) - y^(2) = x^(2)

Does factors to

(h - y)(h + y) = x^(2)

Hope that helps!!!!!

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lenpollock

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2w ago

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