multiplicity

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CP Algebra II 1/7/15
Multiplicity
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The real zeroes of a polynomial correspond to the x-intercepts of the graph of that polynomial. So you can
find the number of real zeroes of a polynomial by looking at the graph, and conversely you can tell how
many times the graph is going to touch or cross the x-axis by looking at the zeroes of the polynomial (or
the factored form of the polynomial).
A zero has a "multiplicity", which refers to the number of times that its associated factor appears in
the polynomial. For instance, the quadratic (x + 3)(x – 2) has the zeroes x = –3 and x = 2, each occuring
4
5
once. The ninth-degree polynomial ( x + 3) ( x - 2 ) (has the same zeroes, but in this case, x = –3 has
multiplicity 4 and x = 2 has multiplicity 5, because of the number of times their factors occur.
The point of multiplicities with respect to graphing is that any factors that occur an even number of times
(twice, four times, six times, etc) are squares, so they don't change sign. Squares are always positive. This
means that the x-intercept corresponding to an even-multiplicity zero can't cross the x-axis, because the
zero can't cause the graph to change sign from positive (above the x-axis) to negative (below the x-axis),
or vice versa. The practical upshot is that an even-multiplicity zero makes the graph just barely touch the
x-a
xis, and then turns it back around the way it came. You can see this in the following graphs:
g(x) = ( x + 3) ( x - 2 )
4
ƒ(x) = (x + 3)(x – 2)
5
As mentioned above both functions have zeros at x = –3 and x = 2. On the second function,
4
5
g(x) = ( x + 3) ( x - 2 ) , the zero at x = – 3 is said to have a multiplicity of 4 and the zero at x = 2 is said
to have multiplicity of 5.
Multiplicity
Passes straight
through:
multiplicity of
one
Shape of a parabola:
multiplicity of even
degree (usually assume
lowest even degree –
2nd degree
General shape is
cubic: multiplicity is
odd greater than or
equal to 3 (usually
assume lowest odd
degree – 3rd degree
Zeros : x = –4 w/ multiplicity of 1, x = –1 with multiplicity of 2, x = 3 with multiplicity of 3
Factors are (x + 4), (x +1), and (x – 3)
A general equation for the polynomial is y = k(x + 4)(x +1)2 (x - 3)3.
If you are given a point on the function, you can substitute for x and y and solve for k.
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