Use the Rational Zero Theorem to find possible rational zeros.
Find zeros of a polynomial function.
Solve polynomial equations.
Use the Linear Factorization Theorem to find polynomials with given zeros.
Apply Descartes’s Rule of Signs.
If a polynomial has integer coefficients and ( \frac{p}{q} ) (where ( \frac{p}{q} ) is in lowest terms) is a rational zero of f:
( p ) is a factor of the constant term.
( q ) is a factor of the leading coefficient.
Goal: List all possible rational zeros.
Constant term: [Insert value].
Leading coefficient: 4.
Factors of the constant term: [List factors].
Factors of the leading coefficient: 1, 2, 4.
Possible rational zeros: [Insert list].
Solution Steps:
List all possible rational zeros.
Apply synthetic division to test possible rational zeros.
After testing, find a rational zero:
Successful rational zero: x = 2.
Find remaining zeros based on the identified rational zero.
Solutions identified are: 2, [Insert other zeros if available].
A polynomial of degree n has n roots (counting multiplicities).
If ( r ) is a root with real coefficients, then its conjugate ( ar{r} ) is also a root.
Imaginary roots occur in conjugate pairs.
List all possible rational roots.
Apply synthetic division.
Through synthetic division, identify x = 1 as a root.
Factor the polynomial accordingly.
Continue to solve using the quadratic formula.
Find the complete solution set of the polynomial equation.
A polynomial of degree n has at least one complex root.
If ( f(x) ) is a polynomial, then it can be expressed as the product of linear factors:
( f(x) = a(x - r_1)(x - r_2)...(x - r_n) ), where ( r_i ) are roots and ( a
eq 0 ).
Find a third-degree polynomial with real coefficients and known zeros, including i.
By the conjugate root theorem, include -i.
Using zeros: [Insert calculations and polynomial formed].
The number of positive real zeros is determined by:
The number of sign changes in ( f(x) ).
Total possibilities: same as sign changes or less by a positive even integer.
The number of negative real zeros is determined by:
The number of sign changes in ( f(-x) ).
Total possibilities: same as sign changes or less by a positive even integer.
Counting sign changes in ( f(x) ): 4 changes.
Possible positive real zeros: 4, 2, or 0.
Counting sign changes in ( f(-x) ): 0 changes.
Conclusion: No negative real roots.