Unit 2: Polynomials — your full class. Work through Day 1 → Day 8. Pass each day's quiz so the Practice Final unlocks.
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To multiply polynomials, distribute every term of the first factor to every term of the second, then combine like terms and write the result in standard form (highest degree first). For two binomials this is FOIL; for longer products, a grid/box keeps track of all the partial products so no term is dropped.
Row n of Pascal's Triangle gives the coefficients of (a + b)^n: row 0 is 1, row 1 is 1 1, row 2 is 1 2 1, row 3 is 1 3 3 1, row 4 is 1 4 6 4 1, and each entry is the sum of the two entries above it. To raise a binomial like (2x - 3)^4, take row n and multiply each entry by descending powers of the first term and ascending powers of the second term, alternating signs.
a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a^3 - b^3 = (a - b)(a^2 + ab + b^2). Remember the signs with SOAP: the binomial keeps the Same sign, the middle term of the trinomial takes the Opposite sign, and the squared term at the end is Always Positive. To spot cubes, check the perfect cubes (1, 8, 27, 64, 125, 216, 1000) and make sure every variable exponent is divisible by 3.
Expand (2x - 3)^4 using Pascal's Triangle.
Answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81
Write a polynomial in standard form that represents the volume of a cylinder with radius (x + 1) and height (x + 4).
Answer: V = pi*x^3 + 6*pi*x^2 + 9*pi*x + 4*pi
Answer: x^3 - 2x^2 - 13x + 6
Answer: x^3 - 12x^2 + 48x - 64
Answer: (x + 4)(x^2 - 4x + 16)
Answer: (3x - 5)(9x^2 + 15x + 25)
1. Which is the correct expansion of (x + 2)^3?
2. Which factorization follows SOAP correctly?
3. Which expression is a sum or difference of cubes in factoring-ready form?
Always pull out the GCF first, then look for difference of squares a^2 - b^2 = (a + b)(a - b), trial-factor trinomials, and four-term grouping. A factorization is complete when every factor is linear or an irreducible quadratic.
A polynomial like ax^4 + bx^2 + c is quadratic IN FORM: substitute u = x^2, factor au^2 + bu + c, then swap u back and keep factoring each piece. Example: x^4 - 3x^2 - 4 = u^2 - 3u - 4 = (u - 4)(u + 1) = (x - 2)(x + 2)(x^2 + 1).
A solution of P(x) = 0 is the same thing as a zero of the polynomial P. Every polynomial of degree n has exactly n complex solutions counting multiplicity, so a 4th-degree equation has four solutions when repeats are counted.
For polynomials with real coefficients, complex zeros come in conjugate pairs: if a + bi is a zero, then a - bi is too; irrational zeros of the form a + b*sqrt(c) come paired with a - b*sqrt(c). Conjugate pairs let you rebuild the polynomial from a partial root list: roots 1/2 and 3i force the factors (2x - 1)(x - 3i)(x + 3i) = (2x - 1)(x^2 + 9).
Factor completely: 4x^4 - 13x^2 + 9.
Answer: (2x - 3)(2x + 3)(x - 1)(x + 1)
Write a degree-3 polynomial equation with rational coefficients that has roots 1/2 and 3i.
Answer: 2x^3 - x^2 + 18x - 9 = 0
Answer: (x + 3)(x - sqrt(2))(x + sqrt(2))
Answer: 2(x - 9)(x + 9)
Answer: (x + 1)(x - 2)(x^2 - x + 1)(x^2 + 2x + 4)
Answer: 3x^3 - 13x^2 - 10x
1. How many solutions does x^4 + 3x^2 - 5 = 0 have over the complex numbers (counting multiplicity)?
2. P(x) has real coefficients and one root is 2 - i. Which other number must also be a root?
3. Which substitution makes f(x) = x^8 - 65x^4 + 64 easiest to factor?
If A*B*C = 0, then at least one factor is 0 -- and conversely, a product of numbers can only be 0 when one of the factors is 0. That is the engine for solving by factoring: factor the polynomial, then set every factor equal to zero.
Rearrange so one side is 0, pull out the GCF, then factor by grouping, difference of squares, trinomial patterns, or a u-substitution -- and never divide both sides by a variable, because that deletes a solution. Solve each linear factor; for a quadratic that will not factor further, use square roots or the quadratic formula.
List the solutions as a set, counting a repeated factor with its multiplicity. Substituting each solution back into the original equation verifies the factoring for free -- a degree-n equation has n solutions once multiplicity is counted.
Solve by factoring: x^3 - 5x^2 - 9x + 45 = 0.
Answer: {3, -3, 5}
Solve by factoring: 2x^2 + 7x - 15 = 0.
Answer: x = 3/2 or x = -5
Answer: x = 3 and x = 5
Answer: x = +/- 2sqrt(2) and x = +/- i*sqrt(3)
Answer: x = 0, 8, or -8
Answer: x = 0, 2, or -4/3
1. Solve x^2 - 5x = 0.
2. List all solutions of (x - 3)^2 (x - 8) = 0.
3. A good FIRST step for solving x^3 - 8x^2 + 15x = 0 is to
Split the numerator into separate fractions and divide each term by the monomial: divide the coefficients and subtract exponents (x^a / x^b = x^(a - b)). Simplify any remaining fractions.
Order both polynomials by descending degree, divide the leading terms, multiply that piece by the divisor, subtract, and bring the next term down -- repeat until the remainder has smaller degree than the divisor. Write the answer as quotient + remainder/(divisor); missing terms get 0 coefficients.
Works only when the divisor is linear, x - c: write c and the dividend's coefficients (0 for missing terms), bring the first coefficient down, then repeatedly multiply by c and add down the columns. The last number is the remainder and the earlier numbers are the quotient's coefficients in descending order; a remainder of 0 means x - c is a factor, which is why synthetic division is the fast way to test candidate zeros.
Divide (2x^3 + 2x^2 + 7x - 8) by (x^2 + 2) using long division.
Answer: (2x + 2) + (3x - 12)/(x^2 + 2)
Divide (x^3 - 8) by (x - 2) using synthetic division.
Answer: x^2 + 2x + 4, remainder 0
Answer: 6x^4 + 4x^2 - 2
Answer: (3x + 4) + (2x + 12)/(x^2 - 1)
Answer: x^2 + 2x - 15, remainder 0
Answer: x^2 - 4, remainder 0
1. (24x^5 - 16x^3) / (4x^2) =
2. To divide f(x) = 2x^3 + x^2 - 5x + 9 by (x - 5) with synthetic division, you should use
3. Dividing f(x) by (x + 3) with synthetic division ends with a final row value of 2, so
Dividing f(x) by the linear factor (x - c) leaves remainder f(c), so synthetic division and direct substitution are interchangeable: the last entry of the synthetic row always equals f(c). This turns every zero question into a quick evaluation.
(x - c) is a factor of f(x) exactly when f(c) = 0, in which case c is called a zero of f. A polynomial g(x) is a factor of f(x) whenever dividing f by g leaves remainder 0 -- test it synthetically or by substitution.
Given one zero c, divide by (x - c) to drop the degree by one, then factor the quotient (trinomials, grouping, difference of squares, or a fresh zero test). Repeat until the polynomial is fully linear or irreducible quadratic; each repeated factor contributes a zero with multiplicity equal to how many times it appears.
Is g(x) = x + 1 a factor of f(x) = x^3 - 7x - 6? If so, factor f completely.
Answer: Yes: (x + 1)(x - 3)(x + 2); zeros x = -1, 3, -2
Find all zeros of f(x) = x^3 - 6x^2 + 9x - 4 given that x = 4 is one zero.
Answer: x = 4 and x = 1 (double root)
Answer: Yes: f(5) = 125 + 250 - 300 - 75 = 0
Answer: x = 2, 3, and -5
Answer: (x + 3)(x + 2)(x - 1)
Answer: x = 1, -1, 2, and -2
1. What is the remainder when f(x) = x^3 + 2x^2 - 5x + 8 is divided by (x - 3)?
2. x + 2 is a factor of x^3 + 8 because
3. Given that x = 4 is a zero of f, the standard next step to find the remaining zeros is to
For a polynomial with integer coefficients, every rational zero p/q written in lowest terms must have p dividing the constant term and q dividing the leading coefficient. The theorem provides a finite candidate list, not a guarantee -- each candidate is then tested by synthetic division (or by evaluating f(p/q)).
If the polynomial is quadratic, factor directly or use the quadratic formula -- no candidate list is needed. When grouping, difference of squares/cubes, or a u-substitution works, factor first; reach for the Rational Zero Theorem only when no algebraic pattern applies and rational zeros are plausible.
A leftover quadratic factor can hold irrational zeros (from the quadratic formula, like (5 +/- sqrt(21))/2) or imaginary zeros (like +/- 2i from x^2 + 4). The conjugate theorems pair these up, and the degree-n count (n zeros total, counting multiplicity) tells you when the list is complete.
List the possible rational zeros of f(x) = 2x^3 - 7x^2 + 9, then find all zeros.
Answer: x = 3, 3/2, and -1
Find all zeros of f(x) = x^3 - 9x^2 + 21x - 4.
Answer: x = 4, (5 + sqrt(21))/2, (5 - sqrt(21))/2
Answer: +/-1, +/-2, +/-4
Answer: x = -1 and x = -3 (double root)
Answer: x = 3, 2i, -2i
Answer: x = 5, -1, -2 + i, -2 - i
1. Which number is NOT a possible rational zero of f(x) = 2x^3 + 3x^2 - 4?
2. A rational zero of f(x) = x^3 - 3x^2 + x - 3 is
3. Synthetic division shows x - 5 is a factor with quotient x^2 - 4x - 5. The remaining zeros are
End behavior is controlled only by the degree and leading (highest-degree) coefficient. Odd degree: the two ends point in opposite directions (positive leading coefficient falls left and rises right; negative rises left and falls right). Even degree: both ends point the same way (positive: both up; negative: both down).
A turning point is where the graph changes from rising to falling or back. A degree-n polynomial has at most n - 1 turning points, and you find the actual count from the graph or by testing neighboring values -- a point higher than both neighbors is a local maximum, lower than both is a local minimum.
At a zero of odd multiplicity (1, 3, ...) the graph crosses the x-axis; at a zero of even multiplicity (2, 4, ...) it just touches and turns around (bounces), because the sign of f does not change across an even-power factor. Multiplicity also flattens the graph near the zero: higher multiplicity means a flatter touch.
Describe the end behavior and maximum number of turning points of g(x) = -2x^4 + x^2 - 3x + 1.
Answer: Both ends fall (g(x) -> -infinity on the far left and far right); at most 3 turning points
Analyze f(x) = (x + 1)^2 (x - 2): end behavior, zero behavior, turning points.
Answer: Falls left, rises right; bounces at x = -1, crosses at x = 2; 2 turning points: (-1, 0) and (1, -4)
Answer: Odd degree, positive leading coefficient: falls left, rises right
Answer: Even degree, negative leading coefficient: both ends fall
Answer: Crosses at x = 0 and x = -1; bounces at x = 3
Answer: Both ends rise; at most 5 turning points
1. As x goes to infinity, f(x) = -x^3 + 4x goes to
2. If the graph of a polynomial touches the x-axis at x = 1 and turns around, the factor (x - 1) has multiplicity
3. What is the maximum possible number of turning points for a degree-7 polynomial?
Build a sketch from four ingredients: (1) x-intercepts = the zeros, with each zero's multiplicity telling you cross (odd) or bounce (even); (2) y-intercept = f(0); (3) end behavior from degree + leading coefficient; (4) a short table of values between and beyond the intercepts to fix how high or low each arc goes. Connect the marked points with a smooth curve.
Evaluate f at integers between the x-intercepts and one or two steps beyond them. The sign pattern (which side of the x-axis each point sits on) must match the cross/bounce rules, and the largest/smallest table values locate the humps and valleys of the sketch.
The degree plus the bounce count tells you the minimum number of turning points (each bounce counts as one), the end behavior tells you how many humps to expect overall (for odd degree: a hump then a valley; for even degree: an extra hump or valley), and every zero must reappear on your curve exactly as its multiplicity demands.
Sketch f(x) = (x + 2)(x - 1)(x - 3): intercepts, end behavior, key points.
Answer: x-int -2, 1, 3 (all crossed); y-int 6; falls left, rises right; key points (-1, 8) and (2, -4)
Sketch f(x) = -x(x - 2)^2: intercepts, end behavior, key points.
Answer: x-int 0 (crossed) and 2 (bounced); y-int 0; rises left, falls right; key points (-1, 9) and (1, -1)
Answer: x-int 0, 2, -2 (all crossed); y-int 0; falls left, rises right; f(-1) = 3, f(1) = -3, f(3) = 15
Answer: x-int -2, -1, 1, 2 (all crossed); y-int 4; both ends rise; f(1/2) = 45/16, f(3/2) = -35/16
Answer: -7 (from f(0))
Answer: x-int -3 (crossed), 2 (bounced); y-int -12; rises left, falls right; f(-4) = 36, f(0) = -12, f(3) = -6
1. The y-intercept of f(x) = x^3 - 2x^2 - 5x + 6 is
2. For f(x) = (x - 2)(x - 4)^2, the graph
3. The end behavior of g(x) = x^4 - 16 is
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