Transcript Warm up - My CCSD

Warm up
Write the quadratic f(x) in vertex form.
Polynomial Functions
Descartes Rule of Signs
Compare the products of:
=
=
Questions:
Do you know that the variation in the signs is f (x). It can
determine the number of positive and negative zeros?
• Descartes’ Rule of Signs
Descartes’ Rule of Signs
• In some cases, the following rule is helpful in
eliminating candidates from lengthy lists of
possible rational roots.
– It was discovered by the French philosopher and
mathematician René Descartes around 1637.
Variation in Sign
• To describe this rule, we need
the concept of variation in sign.
– Suppose P(x) is a polynomial with real coefficients,
written with descending powers of x (and omitting
powers with coefficient 0).
– A variation in sign occurs whenever adjacent
coefficients have opposite signs.
Variation in Sign
• For example,
P(x) = 5x7 – 3x5 – x4 + 2x2 + x – 3
has three variations in sign.
Descartes’ Rule of Signs
• Let P be a polynomial with real coefficients.
1. The number of positive real zeros of P(x) is either
equal to the number of variations in sign in P(x)
or is less than that by an even whole number.
2. The number of negative real zeros of P(x) is either
equal to the number of variations in sign in P(-x)
or is less than that by an even whole number.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x) = x3 + 2x2 + 5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f(x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f(-x). We obtain this equation by replacing x
with -x in the given function.
f(x) = x3 + 2x2 + 5x + 4
Replace x with -x.
f(-x) = (-x)3 + 2(-x)2 + 5(-x) + 4
= -x3 + 2x2 - 5x + 4
This is the given polynomial function.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x) = x3 + 2x2 + 5x + 4.
Solution
Now count the sign changes.
f(-x) = -x3 + 2x2 - 5x + 4
1
2
3
There are three variations in sign. The number of negative real zeros of f is
either equal to the number of sign changes, 3, or is less than this number by
an even integer. This means that there are either 3 negative real zeros or 3 - 2
= 1 negative real zero.
Using Descartes’ Rule
• Use Descartes’ Rule of Signs to determine the
possible number of positive and negative real
zeros of the polynomial
P(x) = 3x6 + 4x5 + 3x3 – x – 3
– The polynomial has one variation in sign.
– So, it has one positive zero.
Using Descartes’ Rule
• Now,
P(–x) = 3(–x)6 + 4(–x)5 +3(–x)3 – (–x) – 3
= 3x6 – 4x5 – 3x3 + x – 3
– Thus, P(–x) has three variations in sign.
– So, P(x) has either three or one negative zero(s),
making a total of either two or four real zeros.
Example 1:
• State the number of positive and negative real
zeros for
Example 1:
• State the number of positive and negative real
zeros for
Your Turn:
• Find the possible number of positive negative
and complex solutions for each of the
following: