1/2 + square root 3/2i
... • Zero: the real zero’s of f(x) are -3(multiplicity 1) and 4( mulitiplicity 2). So f(x) is zero if x=-3 or x=4. • Positive: x^2+1 is positive for all real numbers. (x-4)^2 is positive except x=4. x+3 is positive only if x>-3. So f(x) is positive if x>-3 and x≠4. • Negative: f(x) is negative if x<-3. ...
... • Zero: the real zero’s of f(x) are -3(multiplicity 1) and 4( mulitiplicity 2). So f(x) is zero if x=-3 or x=4. • Positive: x^2+1 is positive for all real numbers. (x-4)^2 is positive except x=4. x+3 is positive only if x>-3. So f(x) is positive if x>-3 and x≠4. • Negative: f(x) is negative if x<-3. ...
Lesson 4
... Polynomial Functions Let’s say I asked you to find all the roots of 10x3+9x2–19x+6=0. How can we solve this? 1. _______ ...
... Polynomial Functions Let’s say I asked you to find all the roots of 10x3+9x2–19x+6=0. How can we solve this? 1. _______ ...
Complex numbers
... The complex numbers C are the set of all numbers of the form a + bi where a, b ∈ R and i2 = −1. We may define addition of complex numbers by (a + bi) + (c + di) = (a + c) + (b + d)i and multiplication as (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i. For any number z = a + bi we ...
... The complex numbers C are the set of all numbers of the form a + bi where a, b ∈ R and i2 = −1. We may define addition of complex numbers by (a + bi) + (c + di) = (a + c) + (b + d)i and multiplication as (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i. For any number z = a + bi we ...
Chapter 4: Polynomials A polynomial is an expression of the form p
... p(X) is a polynomial of degree n, then p(X) cannot have more than n roots. To see this, suppose that p(X) has more than n roots, say a1 , a2 , . . . , am with m > n. Then, according to what we have just learned, f (X) ≡ (X − a1 )(X − a2 ) · · · (X − am ) is a factor of p(X). This cannot happen beca ...
... p(X) is a polynomial of degree n, then p(X) cannot have more than n roots. To see this, suppose that p(X) has more than n roots, say a1 , a2 , . . . , am with m > n. Then, according to what we have just learned, f (X) ≡ (X − a1 )(X − a2 ) · · · (X − am ) is a factor of p(X). This cannot happen beca ...
