algebra ii - MooreMath23
... 2) Always use PLACEHOLDERS for missing coefficients of a descending order polynomial (Example: If x4 – 3x2 + 4x + 6 is given, use 1+ 0 – 3 + 4 + 6 on the first line 3) Remove the variables for this style of division on the first line 4) The Leading Coefficient drops directly to the solution line to ...
... 2) Always use PLACEHOLDERS for missing coefficients of a descending order polynomial (Example: If x4 – 3x2 + 4x + 6 is given, use 1+ 0 – 3 + 4 + 6 on the first line 3) Remove the variables for this style of division on the first line 4) The Leading Coefficient drops directly to the solution line to ...
Lecture Notes for Section 3.3
... If p x is a polynomial with real coefficients, then complex zeros must occur in conjugate pairs. That is, if a bi, b 0 is a zero, then a bi will also be a zero. Corollary III: Number of Zeroes If p x is a polynomial of degree n 1, the p has exactly n zeroes (either real or complex), ...
... If p x is a polynomial with real coefficients, then complex zeros must occur in conjugate pairs. That is, if a bi, b 0 is a zero, then a bi will also be a zero. Corollary III: Number of Zeroes If p x is a polynomial of degree n 1, the p has exactly n zeroes (either real or complex), ...
PreAP PreCal
... A polynomial is an expression of the form: anxn + an-1xn-1+ …+ a2x2 + a1x + a0 Where the a’s are numbers and the n’s are whole numbers. Note that the degree of each term is one smaller than the degree of the previous term. Note that an is the _______________________ and a0 is a __________________. T ...
... A polynomial is an expression of the form: anxn + an-1xn-1+ …+ a2x2 + a1x + a0 Where the a’s are numbers and the n’s are whole numbers. Note that the degree of each term is one smaller than the degree of the previous term. Note that an is the _______________________ and a0 is a __________________. T ...
INTRODUCTION TO ALGEBRA II MIDTERM 1 SOLUTIONS Do as
... Since α is a root of P , we have P (X) = (X − α)S1 (X) for some S1 (X) ∈ C[X]. We have 0 = P (ᾱ) = (ᾱ − α)S1 (ᾱ) and since ᾱ − α 6= 0 (α is not real), ᾱ is a root of S1 , so S1 (X) = (X − ᾱ)S(X) and the results follows. d.– Show that the polynomial (X − α)(X − ᾱ) is in R[X]. Solution: It is e ...
... Since α is a root of P , we have P (X) = (X − α)S1 (X) for some S1 (X) ∈ C[X]. We have 0 = P (ᾱ) = (ᾱ − α)S1 (ᾱ) and since ᾱ − α 6= 0 (α is not real), ᾱ is a root of S1 , so S1 (X) = (X − ᾱ)S(X) and the results follows. d.– Show that the polynomial (X − α)(X − ᾱ) is in R[X]. Solution: It is e ...
