Review the key concepts, formulae, and examples before starting your quiz.
πConcepts
A polynomial function of degree is defined by the expression . Visually, these functions are always smooth and continuous curves with no breaks, holes, or sharp corners (cusps), extending infinitely in both directions along the x-axis.
The end behavior of a polynomial is determined by its degree and leading coefficient . If is even, both ends of the graph point in the same vertical direction (forming a general 'U' or 'W' shape). If is odd, the ends point in opposite directions (forming a general 'S' or 'N' shape). Positive leading coefficients result in the right side of the graph pointing upwards.
The -intercept of the polynomial is the point , where the graph crosses the vertical axis. This is found by evaluating .
Real zeros or roots are the -values where , appearing as -intercepts on the graph. The multiplicity of a root affects the graph's behavior: if is odd, the graph crosses the -axis; if is even, the graph is tangent to the -axis (touches and turns around), creating a local maximum or minimum at that point.
A polynomial of degree can have at most real roots and at most turning points. Turning points are the 'peaks' (local maxima) and 'valleys' (local minima) where the function changes from increasing to decreasing or vice-versa.
The Factor Theorem states that is a factor of if and only if . Geometrically, this means the graph passes exactly through the point . The Remainder Theorem states that the remainder of is equal to the value .
Vieta's Formulas relate the coefficients of a polynomial to sums and products of its roots. For a polynomial of degree , the sum of the roots is and the product of the roots is .
πFormulae
\text{Sum of roots } (\sum \alpha_i) = -\frac{a_{n-1}}{a_n}$
\text{Product of roots } (\prod \alpha_i) = (-1)^n \frac{a_0}{a_n}$
\text{Number of Turning Points} \le n - 1$
π‘Examples
Problem 1:
Find a cubic polynomial that has zeros at , and passes through the point .
Solution:
- Using the zeros, write the factored form: . \ 2. Substitute the point into the equation to find : \ \ \ . \ 3. Write the final equation: . \ 4. (Optional) Expand to standard form: .
Explanation:
Since the zeros are given, the Factor Theorem allows us to construct the function using its linear factors. We include a leading coefficient and solve for it using the provided coordinate point.
Problem 2:
Factorize completely given that is a zero.
Solution:
- Since is a zero, is a factor. \ 2. Perform synthetic division or long division of by : \ Dividing gives the quadratic quotient . \ 3. Factorize the quadratic quotient: \ . \ 4. State the completely factorized form: . \ 5. The zeros are .
Explanation:
We use the Factor Theorem to identify the first linear factor. By dividing the cubic by this factor, we reduce the polynomial to a quadratic, which can then be solved using standard factoring techniques (finding numbers that multiply to and add to ).