Free math worksheets

- The row begins and ends with \(1\). - Each interior entry is the sum of the two entries above it in Pascal's triangle. - A fourth power has five coefficient entries.

1. The coefficient row for the fourth power is \(1,4,6,4,1\). 2. These are the coefficients that multiply the five terms in the expansion of \((a+b)^4\).

\(1,4,6,4,1\)

- How is an entry formed from the two entries above it? - Compare the left and right sides of a row. - What pattern appears at the two ends of every row?

1. Pascal's rule gives \(\binom{14}{7}=\binom{13}{6}+\binom{13}{7}=1716+1716=3432\). 2. The symmetry identity is \(\binom{n}{k}=\binom{n}{n-k}\). Since \(13-6=7\), \(\binom{13}{6}=\binom{13}{7}\). 3. Every row begins and ends with \(1\), so \(\binom{15}{0}=1\) and \(\binom{15}{15}=1\).

a) \(\binom{14}{7}=3432\) b) Since \(13-6=7\), symmetry gives \(\binom{13}{6}=\binom{13}{7}\). c) \(\binom{15}{0}=1\), \(\binom{15}{15}=1\)

- Recall the entries and indexing of row \(n=7\). - Which two values of \(k\) are symmetric in a row? - Replace a large \(k\) with \(n-k\) before calculating. - Compare the number of factors needed for \(\binom{10}{8}\) and \(\binom{10}{2}\).

1. Row \(n=7\) contains \(1,7,21,35,35,21,7,1\). Thus \(35=\binom{7}{3}=\binom{7}{4}\). 2. By symmetry, \(\binom{10}{8}=\binom{10}{2}\). 3. \(\binom{10}{2}=\frac{10\cdot 9}{2}=45\). Using \(k=2\) requires fewer factors than calculating directly with \(k=8\).

a) \(\binom{7}{3}\) and \(\binom{7}{4}\) b) \(\binom{10}{8}=\binom{10}{2}=45\)

- Match each entry in row \(n=5\) with its number of successes, beginning with \(0\). - “More than \(3\)” includes \(4\) and \(5\). - With equally likely paths, divide favorable paths by all paths.

1. The entry for exactly \(3\) successes is \(\binom{5}{3}=10\). 2. More than \(3\) successes means \(4\) or \(5\) successes, so the number of paths is \(\binom{5}{4}+\binom{5}{5}=5+1=6\). 3. Exactly one success has \(\binom{5}{1}=5\) equally likely paths out of \(32\), so the probability is \(\frac{5}{32}=0.15625\).

a) \(10\) b) \(6\) c) \(\frac{5}{32}=0.15625\)

- Compare the coefficients with the expansion of \((a + b)^3\). - A power of a linear factor equals zero only when the linear factor equals zero. - The exponent on the factor gives the multiplicity.

1. The identity \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) with \(a = x\) and \(b = 1\) gives \(f(x) = (x + 1)^3\). 2. Set \((x + 1)^3 = 0\). Then \(x + 1 = 0\), so \(x = -1\). 3. Because the factor \(x + 1\) occurs to the third power, the zero has multiplicity 3.

1. \(f(x) = (x + 1)^3\) 2. \(x = -1\) with multiplicity 3

- Use the Pascal row associated with a fourth power. - The exponent of \(x\) decreases from \(4\) to \(0\) while the exponent of \(y\) increases from \(0\) to \(4\). - Each term has total exponent \(4\).

1. The coefficient row for the fourth power is \(1,4,6,4,1\). 2. Pair these coefficients with descending powers of \(x\) and ascending powers of \(y\). 3. The expansion is \(x^4+4x^3y+6x^2y^2+4xy^3+y^4\).

\(x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

- Start with the same coefficient row as for \((x+y)^4\). - Remember that the second binomial term is \(-y\), not \(y\). - Track whether each power of \(-y\) is positive or negative.

1. Use coefficients \(1,4,6,4,1\). 2. Treat the second term as \(-y\). Its powers alternate the signs of successive terms. 3. The expansion is \(x^4-4x^3y+6x^2y^2-4xy^3+y^4\).

\(x^4-4x^3y+6x^2y^2-4xy^3+y^4\)

- Determine how many times the second binomial term must be selected to produce \(y^2\). - Use a binomial coefficient to count the choices of those factor positions. - Include the numerical coefficient carried by each selected \(2y\) factor.

1. To obtain \(x^3y^2\), choose the \(2y\) term from exactly \(2\) of the \(5\) factors. 2. The corresponding term is \(\binom{5}{2}x^3(2y)^2\). 3. Its coefficient is \(10\cdot4=40\).

\(40\)

- Think of the expansion as making one choice from each of five factors. - Count how many factors contribute \(a\) and how many contribute \(b\). - Compare choosing positions for \(b\) with choosing the complementary positions for \(a\).

1. Each term in the expansion comes from choosing either \(a\) or \(b\) from each of the \(5\) identical factors. 2. If \(b\) is chosen from \(k\) factors, then \(a\) is chosen from the remaining \(5-k\) factors. The term therefore has exponents whose sum is \((5-k)+k=5\). 3. The coefficient is \(\binom{5}{k}\). Choosing the \(k\) positions for \(b\) is equivalent to choosing the \(5-k\) positions for \(a\), so \(\binom{5}{k}=\binom{5}{5-k}\). This gives symmetric coefficients.

Each of the \(5\) factors contributes either one \(a\) or one \(b\), so the exponents total \(5\). The coefficients are symmetric because \(\binom{5}{k}=\binom{5}{5-k}\).

- Add adjacent entries to form the next row. - Use the recurrence formula or the definition of a binomial coefficient. - Add the entries of row \(n=5\) and compare the result with \(2^5\).

1. Add adjacent entries from the given row: the next row begins \(1\), \(1+9=10\), \(9+36=45\), \(36+84=120\), \(84+126=210\), and \(126+126=252\). 2. \(\binom{10}{4}=120\cdot\frac{10-3}{4}=120\cdot\frac{7}{4}=210\). 3. Row \(n=5\) is \(1,5,10,10,5,1\). Its sum is \(1+5+10+10+5+1=32=2^5\).

a) \(1,10,45,120,210,252\) b) \(\binom{10}{4}=210\) c) \(1+5+10+10+5+1=32=2^5\)

- Form the next Pascal's triangle row by adding adjacent entries. - Match each coefficient with its number of correct answers. - “At least \(5\)” includes \(5\) and \(6\). - How many total answer patterns are possible for \(6\) two-choice questions?

1. Row \(n=6\) is \(1,6,15,20,15,6,1\). 2. Exactly \(3\) correct answers can occur in \(\binom{6}{3}=20\) answer patterns. 3. At least \(5\) correct answers means \(5\) or \(6\). The number of favorable patterns is \(\binom{6}{5}+\binom{6}{6}=6+1=7\). There are \(2^6=64\) equally likely patterns, so the probability is \(\frac{7}{64}=0.109375\).

a) \(1,6,15,20,15,6,1\) b) \(20\) c) \(\frac{7}{64}=0.109375\)

- Reversing the order in an odd power changes the sign: compare \((a - b)^3\) and \((b - a)^3\). - Use cubic binomial identities or expand the factors directly. - Read multiplicity from the exponent on a linear factor. - A quadratic factor with no real zeros does not add real zeros or change the multiplicity of an existing linear factor.

1. At \(x = 3\), all five functions equal \(0\). At \(x = 0\), the values are \(f_1(0) = -27\), \(f_2(0) = -27\), \(f_3(0) = 27\), \(f_4(0) = -27\), and \(f_5(0) = -27\). 2. Function \(f_1\) has the zero \(x = 3\) with multiplicity 3. Function \(f_3(x) = -(x - 3)^3\) has the same zero and multiplicity. Expanding \(f_5\) gives \((x - 3)^3\), so \(f_5\) also has \(x = 3\) with multiplicity 3. 3. Factor the difference of cubes: \(f_2(x) = x^3 - 27 = (x - 3)(x^2 + 3x + 9)\). The quadratic factor has no real zeros, so \(f_2\) has \(x = 3\) with multiplicity 1. Function \(f_4\) is already the same product and has the same zero and multiplicity. 4. The binomial expansion gives \((x - 3)^3 = x^3 - 9x^2 + 27x - 27\), so \(f_1 = f_5\). The difference-of-cubes identity gives \(f_2 = f_4\). Function \(f_3 = -f_1\), so it is not identical to any other listed function.

a) At \(x = 3\), all five values are \(0\). At \(x = 0\): \(f_1 = -27\), \(f_2 = -27\), \(f_3 = 27\), \(f_4 = -27\), \(f_5 = -27\). b) \(f_1\), \(f_3\), and \(f_5\): \(x = 3\) with multiplicity 3 \(f_2\) and \(f_4\): \(x = 3\) with multiplicity 1 c) \(f_1 = f_5\) and \(f_2 = f_4\).

- List the sequences with exactly two makes and one miss. - Count how many such sequences there are. - Compare that count with the coefficients in the expansion of \((p+q)^3\).

1. The probability of a make is \(p=0.8\), and the probability of a miss is \(q=1-0.8=0.2\). 2. Exactly two makes occur in the sequences MMX, MXM, and XMM, where M represents a make and X represents a miss. 3. Each sequence has probability \(p^2q=(0.8)^2\cdot 0.2=0.128\). Therefore, \(P(\text{exactly two makes})=3p^2q=3\cdot(0.8)^2\cdot 0.2=0.384\). 4. The binomial expansion is \((p+q)^3=p^3+3p^2q+3pq^2+q^3\). The term \(3p^2q\) represents exactly two made shots and one miss.

a) \(0.384=38.4\%\) b) The term is \(3p^2q\).

- Record the letter sequence on each complete path. - Focus on paths containing two \(a\) choices and one \(b\) choice. - Recall what \(\binom{n}{k}\) counts.

a) A three-stage tree with choices \(a\) and \(b\) has \(2^3=8\) paths. A path producing \(a^2b\) must contain two \(a\) factors and one \(b\) factor, so there are \(3\) such paths. b) The paths are \((a,a,b)\), \((a,b,a)\), and \((b,a,a)\). c) The coefficient \(\binom{3}{2}=3\) counts the ways to choose the two of the three positions occupied by \(a\). This is exactly the number of paths producing \(a^2b\).

a) \(3\) paths b) \(a\to a\to b\), \(a\to b\to a\), and \(b\to a\to a\) c) \(\binom{3}{2}=3\) counts the ways to place two \(a\) factors among three positions.

- Use the fourth-power coefficient row, but do not forget that the first binomial term is \(2x\). - Apply each exponent to the entire term \(2x\). - Simplify numerical powers after setting up all five terms.

1. Use coefficients \(1,4,6,4,1\). 2. Substitute \(2x\) for the first binomial term: \((2x)^4+4(2x)^3y+6(2x)^2y^2+4(2x)y^3+y^4\). 3. Simplify to obtain \(16x^4+32x^3y+24x^2y^2+8xy^3+y^4\).

\(16x^4+32x^3y+24x^2y^2+8xy^3+y^4\)

- Use the Pascal coefficients for a fifth power. - Keep \(-2\) inside parentheses when raising it to powers. - Check that the signs alternate because the second binomial term is negative.

1. Use the fifth-power coefficients \(1,5,10,10,5,1\). 2. Treat the second term as \(-2\): \(x^5+5x^4(-2)+10x^3(-2)^2+10x^2(-2)^3+5x(-2)^4+(-2)^5\). 3. Simplify to get \(x^5-10x^4+40x^3-80x^2+80x-32\).

\(x^5-10x^4+40x^3-80x^2+80x-32\)

- Compare the coefficient pattern with a Pascal row for a fourth power. - The alternating signs suggest a negative second binomial term. - Use the constant term to infer the magnitude of that second term.

1. A fourth-power binomial uses Pascal coefficients \(1,4,6,4,1\). 2. The expression \((x-2)^4\) expands with terms \(x^4\), \(4x^3(-2)\), \(6x^2(-2)^2\), \(4x(-2)^3\), and \((-2)^4\). 3. These simplify to \(x^4-8x^3+24x^2-32x+16\), matching the given polynomial.

\((x-2)^4\)

- Keep the second binomial term as \(-b\) while considering its powers. - Compare what happens to even and odd powers of a negative quantity. - The Pascal coefficients themselves do not carry the alternating signs.

1. The coefficients \(1,4,6,4,1\) are correct, but the signs must come from powers of \(-b\). 2. Even powers of \(-b\) are positive and odd powers are negative. 3. Therefore, the middle term \(6a^2b^2\) must be positive. 4. The correct expansion is \(a^4-4a^3b+6a^2b^2-4ab^3+b^4\).

The \(a^2b^2\) term has the wrong sign. The correct expansion is \(a^4-4a^3b+6a^2b^2-4ab^3+b^4\).

- Follow complete root-to-leaf paths and record the four edge labels in order. - Keep only paths with two occurrences of each letter. - Relate the path count to choosing which positions contain one of the letters.

a) There are \(6\) such paths. b) The sequences are \(aabb\), \(abab\), \(abba\), \(baab\), \(baba\), and \(bbaa\). c) Choosing which \(2\) of the \(4\) positions contain \(a\) determines the path. There are \(\binom{4}{2}=6\) choices, so the coefficient of \(a^2b^2\) is \(6\).

a) \(6\) b) \(aabb\), \(abab\), \(abba\), \(baab\), \(baba\), \(bbaa\) c) The count is \(\binom{4}{2}=6\), which is the coefficient of \(a^2b^2\).

- First determine how many factors must contribute the \(2x\) term to produce \(x^3\). - Use a binomial coefficient for the number of position choices. - Include both the power of \(2\) and the power of \(k\) in the coefficient equation.

1. An \(x^3\) term occurs when \(2x\) is chosen from \(3\) of the \(6\) factors and \(k\) is chosen from the other \(3\). 2. The coefficient of \(x^3\) is \(\binom{6}{3}2^3k^3=20\cdot8\cdot k^3=160k^3\). 3. Set \(160k^3=4320\). Then \(k^3=27\), so \(k=3\) because \(k>0\).

\(k=3\)