Catégories : Tous - inequalities - theorem - polynomials

par Fuad Jahen Il y a 13 années

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Polynomial Equations and Inequalities

Polynomial functions and their properties are crucial in algebra. The remainder theorem simplifies the process of finding the remainder when a polynomial \( P(x) \) is divided by \(

Polynomial Equations and Inequalities

Polynomial Equations and Inequalities

Polynomial Equations

Factor Theorem

(x-b) is a factor of a polynomial P(X) if and only if P(b) = 0. Similarly, (ax-b) is a factor of P(x) if and only if P(b/a) = 0 Example: To determine if x-3 is a factor of P(x)=x³ - x² - 14x + 24, calculate P(3): P(3) = (3)³- (3)² + 14(3) + 24 = 27 – 9 – 42 + 24
Rational Zero Theorem

Suppose P(x) is a polynomial function with integer coefficients and x= b/a is a zero of P(x), where a and b are integer and a≠0. Then, b is a factor of the constant term of P(x) a is a factor of the leading coefficient of P(x) ax-b is a factor of P(x) Example: Consider a factorable polynomial such as P(x) = 3x² + 7x – 2. Since the leading coefficient is 3, one of the factors must be of the form (3x-b), where b is a factor of the constant term 2 and P(b/3) = 0 To determine the values of x that should be tested to find b, the integral zero theorem is extended to include polynomials with leading coefficients that are not one.

Integral Zero Theorem

If (x-b) is a factor of a polynomial function P(x) with leading coefficient 1 and remaining coefficients that are integers, then b is a factor of the constant term of P(x). Example: Consider the polynomial P(x) = x² + 6x – 5 A value that satisfies P(b) = 0 also satisfies b² + 6b – 5 =0, or b² + 6b = 5. Since the product of b² + 6b – 5 is 5, the possible integer values for the factors in the product are the factors of 5. They are ±1,±5.

Remainder Theorem

The remainder theorem can also be used to find the point (x,y)
P(x)/(x-b)=Q(x)+R/(x-b) Q(x)=P(x)/(x-b)-R/(x-b) Q(x)=(P(x)-R)/(x-b) Therefore: [x,P(x) ],(b,R)
The remainder theorem states that when a polynomial function P(x) is divided by x-b, the remainder is P(b); and when it is divided by ax-b, the remainder is P(b/a), where a and b are integers, and a≠0
[Q(x) ](ax-b)+R=P(x) [Q(b/a) ][a(b/a)-b]+R=P(b/a) Q(b/a)(0)+R=P(b/a) R=P(b/a)
[Q(x) ](x-b)+R=P(x) [Q(b) ](b-b)+R=P(b) R=P(b)

Families of Polynomial Function

Solving Inequalities

Pre-requisite Skills

[Q(x) ](x-b)+R=P(x)

Corresponding Statement

(P(x))/(x-b)=Q(x)+R/(x-b) ,x≠b

Quotient Form

Q(x) (x-b)√(P(x) ¦_ R

LONG DIVISION FORM

Q(x) = Quotient

P(x) = Polynomial equation/ Dividend

R = Remainder

(x-b) = Divisor/Binomial equation