Algebra 2 Homework Notes

Algebra 2 Homework Notes
Section 3.6: Graphing Polynomial Functions

Factored Form and Standard Form

In this section, students are required to write given polynomials in factored form and then convert them into standard form. Understanding both forms is crucial for analyzing and graphing polynomial functions.

  • Factored Form facilitates the identification of the polynomial's zeros and their multiplicities, which directly relate to the x-intercepts of the graph. For instance, a polynomial in factored form P(x)=a(x−r<em>1)m</em>1(x−r<em>2)m</em>2…P(x) = a(x-r<em>1)^{m</em>1}(x-r<em>2)^{m</em>2}… clearly shows its zeros at x=r<em>1x=r<em>1, x=r</em>2x=r</em>2, etc.

  • Standard Form is useful for determining the degree of the polynomial, its leading coefficient, and its end behavior. A polynomial in standard form is written as P(x)=a<em>nxn+a</em>n−1xn−1+…+a<em>1x+a</em>0P(x) = a<em>n x^n + a</em>{n-1} x^{n-1} + … + a<em>1 x + a</em>0, where ana_n is the leading coefficient and nn is the degree.

It's essential to also identify key features of the polynomial functions, specifically:

Key Features of Polynomials

  • Zeros: These are the x-values for which the polynomial function equals zero, i.e., P(x)=0P(x)=0. Graphically, zeros correspond to the x-intercepts where the graph crosses or touches the x-axis. They are found by setting each factor in the factored form to zero and solving for the variable. For example, if (x−a)(x-a) is a factor, then x=ax=a is a zero.

  • Multiplicities: The multiplicity of a zero is the number of times its corresponding factor appears in the factored form of the polynomial.

    • A zero with an odd multiplicity (e.g., 1, 3, 5…) means the graph crosses the x-axis at that point. A multiplicity of 1 typically means a straight crossing, while higher odd multiplicities (e.g., 3) result in the graph flattening out as it crosses the x-axis, resembling an 'S' shape.

    • A zero with an even multiplicity (e.g., 2, 4, 6…) means the graph touches the x-axis at that point and then turns around (bounces off) without crossing it. A multiplicity of 2 causes the graph to resemble a parabola's vertex at the x-axis.

  • Factored Form: The structure of the polynomial written as a product of its linear and irreducible quadratic factors. This form is derived by factoring the standard form polynomial and is particularly useful for quickly identifying the zeros and their multiplicities.

  • Standard Form: The polynomial expressed as a sum of its terms, typically arranged in descending order of their degrees. This form, a<em>nxn+a</em>n−1xn−1+…+a<em>1x+a</em>0a<em>n x^n + a</em>{n-1} x^{n-1} + … + a<em>1 x + a</em>0, clearly shows the leading term (a<em>nxna<em>n x^n), which dictates the polynomial's end behavior, and the constant term (a</em>0a</em>0), which is the y-intercept.

The students must fill in the respective sections for each polynomial, documenting:

  • Zeros (the x-intercepts)

  • Multiplicities (how the graph behaves at each zero)

  • Factored Form (the product of factors)

  • Standard Form (the expanded polynomial)

Multiplication of Polynomials

In this section, students are tasked to multiply several polynomial expressions and leave them in standard form. This involves applying the distributive property repeatedly or using systematic methods like FOIL (for binomials) or the tabular method.

Level #1 Problems

These problems typically involve distributing a monomial to a binomial.

  1. Multiply: 7(v−7)7(v - 7)

    • Method: Apply the distributive property: multiply 77 by each term inside the parentheses.

    • Steps: 7⋅v−7⋅77 \cdot v - 7 \cdot 7

    • Standard Form Result: 7v−497v - 49

Level #2 Problems

These problems involve distributing monomials to binomials or multiplying two binomials (FOIL method) or a binomial by a trinomial.

  1. Multiply: 5v(7v−2)5v(7v - 2)

    • Method: Distribute the monomial 5v5v to each term in the binomial.

    • Steps: (5v⋅7v)−(5v⋅2)(5v \cdot 7v) - (5v \cdot 2)

    • Standard Form Result: 35v2−10v35v^2 - 10v

  2. Multiply: 8(5n−1)8(5n - 1)

    • Method: Distribute the constant 88 to each term.

    • Standard Form Result: 40n−840n - 8

  3. Multiply: 4(7k+8)4(7k + 8)

    • Method: Distribute the constant 44 to each term.

    • Standard Form Result: 28k+3228k + 32

  4. Multiply: (v+3)(4v+4)(v + 3)(4v + 4)

    • Method: Use the FOIL method (First, Outer, Inner, Last) or the distributive property twice.

      • First: v⋅4v=4v2v \cdot 4v = 4v^2

      • Outer: v⋅4=4vv \cdot 4 = 4v

      • Inner: 3⋅4v=12v3 \cdot 4v = 12v

      • Last: 3⋅4=123 \cdot 4 = 12

    • Steps: 4v2+4v+12v+124v^2 + 4v + 12v + 12

    • Standard Form Result: 4v2+16v+124v^2 + 16v + 12

  5. Multiply: (2p−2)(4p−3)(2p - 2)(4p - 3)

    • Method: Apply the FOIL method.

    • Steps: (2p⋅4p)+(2p⋅−3)+(−2⋅4p)+(−2⋅−3)=8p2−6p−8p+6(2p \cdot 4p) + (2p \cdot -3) + (-2 \cdot 4p) + (-2 \cdot -3) = 8p^2 - 6p - 8p + 6

    • Standard Form Result: 8p2−14p+68p^2 - 14p + 6

  6. Multiply: (7n2+3n−7)(n+4)(7n^2 + 3n - 7)(n + 4)

    • Method: Distribute each term of the trinomial to each term of the binomial, or use a tabular method.

    • Steps:

      • 7n2(n+4)+3n(n+4)−7(n+4)7n^2(n+4) + 3n(n+4) - 7(n+4)

      • (7n3+28n2)+(3n2+12n)−(7n+28)(7n^3 + 28n^2) + (3n^2 + 12n) - (7n + 28)

      • Combine like terms: 7n3+(28n2+3n2)+(12n−7n)−287n^3 + (28n^2 + 3n^2) + (12n - 7n) - 28

    • Standard Form Result: 7n3+31n2+5n−287n^3 + 31n^2 + 5n - 28

  7. Multiply: (x2+3x−5)(8x−6)(x^2 + 3x - 5)(8x - 6)

    • Method: Similar to problem 7, distribute each term of the trinomial to each term of the binomial.

    • Steps:

      • x2(8x−6)+3x(8x−6)−5(8x−6)x^2(8x-6) + 3x(8x-6) - 5(8x-6)

      • (8x3−6x2)+(24x2−18x)−(40x−30)(8x^3 - 6x^2) + (24x^2 - 18x) - (40x - 30)

      • Combine like terms: 8x3+(−6x2+24x2)+(−18x−40x)−(−30)8x^3 + (-6x^2 + 24x^2) + (-18x - 40x) - (-30)

    • Standard Form Result: 8x3+18x2−58x+308x^3 + 18x^2 - 58x + 30

  8. Multiply: (x2+x+6)(x2−3)(x^2 + x + 6)(x^2 - 3)

    • Method: Distribute each term of the first trinomial to the second binomial.

    • Steps:

      • x2(x2−3)+x(x2−3)+6(x2−3)x^2(x^2 - 3) + x(x^2 - 3) + 6(x^2 - 3)

      • (x4−3x2)+(x3−3x)+(6x2−18)(x^4 - 3x^2) + (x^3 - 3x) + (6x^2 - 18)

      • Combine like terms: x4+x3+(−3x2+6x2)−3x−18x^4 + x^3 + (-3x^2 + 6x^2) - 3x - 18

    • Standard Form Result: x4+x3+3x2−3x−18x^4 + x^3 + 3x^2 - 3x - 18

  9. Multiply: (p2−7p+5)(6p+2)(p^2 - 7p + 5)(6p + 2)

    • Method: Distribute each term of the trinomial to each term of the binomial.

    • Steps:

      • p2(6p+2)−7p(6p+2)+5(6p+2)p^2(6p+2) - 7p(6p+2) + 5(6p+2)

      • (6p3+2p2)−(42p2+14p)+(30p+10)(6p^3 + 2p^2) - (42p^2 + 14p) + (30p + 10)

      • Combine like terms: 6p3+(2p2−42p2)+(−14p+30p)+106p^3 + (2p^2 - 42p^2) + (-14p + 30p) + 10

    • Standard Form Result: 6p3−40p2+16p+106p^3 - 40p^2 + 16p + 10

Level #3 Problems

These problems typically involve multiplying two trinomials, requiring more extensive distribution and careful combination of like terms. A tabular or vertical multiplication method can be very helpful here to organize terms.

  1. Multiply: (4a2+6a−6)(4a2−2a−5)(4a^2 + 6a - 6)(4a^2 - 2a - 5)

    • Method: Distribute each term of the first trinomial to every term of the second trinomial.

    • Steps:

      • 4a2(4a2−2a−5)+6a(4a2−2a−5)−6(4a2−2a−5)4a^2(4a^2 - 2a - 5) + 6a(4a^2 - 2a - 5) - 6(4a^2 - 2a - 5)

      • (16a4−8a3−20a2)+(24a3−12a2−30a)−(24a2−12a−30)(16a^4 - 8a^3 - 20a^2) + (24a^3 - 12a^2 - 30a) - (24a^2 - 12a - 30)

      • Combine like terms:

      • a4a^4 term: 16a416a^4

      • a3a^3 terms: −8a3+24a3=16a3-8a^3 + 24a^3 = 16a^3

      • a2a^2 terms: −20a2−12a2−24a2=−56a2-20a^2 - 12a^2 - 24a^2 = -56a^2

      • aa terms: −30a+12a=−18a-30a + 12a = -18a

      • Constant term: +30+30

    • Standard Form Result: 16a4+16a3−56a2−18a+3016a^4 + 16a^3 - 56a^2 - 18a + 30

  2. Multiply: (6n2+n+2)(5n2+n+7)(6n^2 + n + 2)(5n^2 + n + 7)

    • Method: Similar to problem 11, distribute each term of the first trinomial to every term of the second trinomial.

    • Steps:

      • 6n2(5n2+n+7)+n(5n2+n+7)+2(5n2+n+7)6n^2(5n^2 + n + 7) + n(5n^2 + n + 7) + 2(5n^2 + n + 7)

      • (30n4+6n3+42n2)+(5n3+n2+7n)+(10n2+2n+14)(30n^4 + 6n^3 + 42n^2) + (5n^3 + n^2 + 7n) + (10n^2 + 2n + 14)

      • Combine like terms:

      • n4n^4 term: 30n430n^4

      • n3n^3 terms: 6n3+5n3=11n36n^3 + 5n^3 = 11n^3

      • n2n^2 terms: 42n2+n2+10n2=53n242n^2 + n^2 + 10n^2 = 53n^2

      • nn terms: 7n+2n=9n7n + 2n = 9n

      • Constant term: 1414

    • Standard Form Result: 30n4+11n3+53n2+9n+1430n^4 + 11n^3 + 53n^2 + 9n + 14

Conclusion

This homework focuses on the essential operations with polynomials, emphasizing the transition between factored forms and standard forms while assessing critical features such as zeros and multiplicities. Students should practice these types of problems to become proficient in handling polynomial expressions both in theoretical and practical applications. Mastery of these concepts is fundamental for further studies in algebra and calculus, as they provide the building blocks for understanding more complex functions and their behaviors.