Question

ap precalculus worksheet 1.11
name: period:
row 0 → 1
row 1 → 1 1
row 2 → 1 2 1
row 3 → 1 3 3 1
1 4 6 4 1
1 5 (10) 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
in pascals triangle (pictured), the 5ᵗʰ row is circled.
in the 5ᵗʰ row, the (first) number 10 has also been circled.
the circled 10 is the 2ⁿᵈ element of row 5... so (5 2)=10.
the notation (n r) represents a “combination”, where (n r)=n!/r!(n - r)!
you will not need to know this formula for the ap precalculus exam!
additional notation: (n r)=ₙcᵣ = c(n,r)

1. deduce the pattern within pascals triangle and add row 8 to the image above.
2. use pascals triangle to expand (x + 2)⁵.
3. use pascals triangle to expand (2x - 1)⁴.
4. use pascals triangle to expand (3x + 4)⁶.
5. use pascals triangle to expand (x² + 1)³.
6. what is the coefficient of the term containing x⁴ when the expression (x + 5)⁶ is expanded?
7. what is the coefficient of the term containing x³ when the expression (x - 3)⁸ is expanded?

1.11 equivalent representations of polys and rats
created by bryan passwater. modified by jamie gardner

Explanation:

Step1: Recall Pascal's Triangle pattern

Each row starts and ends with 1, and each interior number is the sum of the two numbers directly above it.

Step2: Write out Row 8 of Pascal's Triangle

Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 56 28 8 1

Step3: Expand \((x + 2)^5\) using Pascal's Triangle

The coefficients from the 5 - th row (1 5 10 10 5 1) are used.
\((x + 2)^5=1\times x^5\times2^0+5\times x^4\times2^1 + 10\times x^3\times2^2+10\times x^2\times2^3+5\times x^1\times2^4+1\times x^0\times2^5\)
\(=x^5 + 10x^4+40x^3 + 80x^2+80x + 32\)

Step4: Expand \((2x-1)^4\) using Pascal's Triangle

The coefficients from the 4 - th row (1 4 6 4 1) are used.
\((2x-1)^4=1\times(2x)^4\times(-1)^0+4\times(2x)^3\times(-1)^1+6\times(2x)^2\times(-1)^2+4\times(2x)^1\times(-1)^3+1\times(2x)^0\times(-1)^4\)
\(=16x^4-32x^3 + 24x^2-8x + 1\)

Step5: Expand \((3x + 4)^6\) using Pascal's Triangle

The coefficients from the 6 - th row (1 6 15 20 15 6 1) are used.
\((3x+4)^6=1\times(3x)^6\times4^0+6\times(3x)^5\times4^1+15\times(3x)^4\times4^2+20\times(3x)^3\times4^3+15\times(3x)^2\times4^4+6\times(3x)^1\times4^5+1\times(3x)^0\times4^6\)
\(=729x^6+5832x^5+23328x^4+497664x^3+663552x^2+5505024x + 4096\)

Step6: Expand \((x^2 + 1)^3\) using Pascal's Triangle

The coefficients from the 3 - th row (1 3 3 1) are used.
\((x^2+1)^3=1\times(x^2)^3\times1^0+3\times(x^2)^2\times1^1+3\times(x^2)^1\times1^2+1\times(x^2)^0\times1^3\)
\(=x^6+3x^4+3x^2 + 1\)

Step7: Find coefficient of \(x^4\) in \((x + 5)^6\)

The coefficients from the 6 - th row (1 6 15 20 15 6 1) are used. The term with \(x^4\) is \(15\times x^4\times5^2\), so the coefficient is \(15\times25 = 375\)

Step8: Find coefficient of \(x^3\) in \((x - 3)^8\)

The coefficients from the 8 - th row (1 8 28 56 70 56 28 8 1) are used. The term with \(x^3\) is \(56\times x^3\times(-3)^5\), so the coefficient is \(56\times(-243)=-13608\)

Answer:

Row 8 of Pascal's Triangle: 1 8 28 56 70 56 28 8 1
\((x + 2)^5=x^5 + 10x^4+40x^3 + 80x^2+80x + 32\)
\((2x-1)^4=16x^4-32x^3 + 24x^2-8x + 1\)
\((3x+4)^6=729x^6+5832x^5+23328x^4+497664x^3+663552x^2+5505024x + 4096\)
\((x^2+1)^3=x^6+3x^4+3x^2 + 1\)
Coefficient of \(x^4\) in \((x + 5)^6\) is 375
Coefficient of \(x^3\) in \((x - 3)^8\) is - 13608