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!!!!!
d
14c^(2)d &(+) 42c^(3)d This factors to 14c^(2)d(1 + 3c) Hence the GCF is 14c^(2)d Method We have '14' & '3 X 14 = 42'. So '14' is a common factor We have c^(2) = c X c & c^(3) = c X c X c . Sp c^(2) is a common factor. Finally we have 'd' & 'd' . So 'd' is the final common factor. Combining 14 x c^(2) X d = 14c^(2)d as the GCF .
d2/(d - c) + c2/(c - d) = d2/(d - c) - c2/(d - c) = (d2 - c2)/(d - c) = (d + c)(d - c)/(d - c) = d + c
You can't.If f: D --> C where D is the domain of the function f and C is its codomain and D = Ø, then there are no d Є D. Therefore there are no c Є C : f(d) = c. Thus there are no ordered pairs (d, c) to graph.
It is useful to know the linear factors of a polynomial because they give you the zeros of the polynomial. If (x-c) is one of the linear factors of a polynomial, then p(c)=0. Here the notation p(x) is used to denoted a polynomial function at p(c) means the value of that function when evaluated at c. Conversely, if d is a zero of the polynomial, then (x-d) is a factor.
Suppose you have a fraction in the form a/b and suppose c is a common factor of a and b.c is a factor of a so that a = c*xc is a factor of b so that b = c*ywhere x and y are integers.And so a/b = cx/cy = x/y.The process is as follows:find a common factor, c, of the numerator (a) and the denominator (b).the new numerator is the old numerator divided by the common factor that is, x = a/c;the new denominator is the old denominator divided by the common factor that is, y = b/c;the new fraction is x/y.
(c + d)(c - d)
(c + d)(c + d)
14c^(2)d &(+) 42c^(3)d This factors to 14c^(2)d(1 + 3c) Hence the GCF is 14c^(2)d Method We have '14' & '3 X 14 = 42'. So '14' is a common factor We have c^(2) = c X c & c^(3) = c X c X c . Sp c^(2) is a common factor. Finally we have 'd' & 'd' . So 'd' is the final common factor. Combining 14 x c^(2) X d = 14c^(2)d as the GCF .
a(c+d)+b(c+d)=(a+b)(c+d)
yes, 5(c-d) this means 5 times c-d
0
There is a formula for the difference of squares. In this case, the answer is (C + D)(C - D)
D is the choice that is not true.
a b c d e f g
8 is a factor of 40
d. 3
3 x 5 x c x d = 15cd