Question

date period hw #6 - polynomial division: synthetic (2.7) 1. use synthetic division to complete the following problems. a) (5x³ - 9x² - 18x - 1) ÷ (x - 3) b) (11x⁴ + 12x³ + 7x² + 11x + 12) ÷ (x + 1) 2 determine if d(x) is a factor of f(x). a) f(x) = x³ - 5x² - 7x - 2 d(x) = x + 1 b) f(x) = 2x⁴ + x³ - 19x² - 17x - 15 d(x) = x + 3

Explanation:

Step1: Recall Synthetic Division Steps

For a polynomial \( f(x) = a_nx^n + \dots + a_1x + a_0 \) divided by \( (x - c) \), synthetic division uses the coefficients \( a_n, \dots, a_1, a_0 \) and \( c \).

Part 1a: \( (5x^3 - 9x^2 - 18x - 1) \div (x - 3) \)

Step2: Set Up Synthetic Division

Coefficients of dividend: \( 5, -9, -18, -1 \). \( c = 3 \) (since divisor is \( x - 3 \)).
Bring down the first coefficient: \( 5 \).

Step3: Multiply and Add

• Multiply \( 5 \times 3 = 15 \). Add to next coefficient: \( -9 + 15 = 6 \).
• Multiply \( 6 \times 3 = 18 \). Add to next coefficient: \( -18 + 18 = 0 \).
• Multiply \( 0 \times 3 = 0 \). Add to last coefficient: \( -1 + 0 = -1 \).

The result is \( 5x^2 + 6x + 0 - \frac{1}{x - 3} \), so quotient \( 5x^2 + 6x \), remainder \( -1 \).

Part 1b: \( (11x^4 + 12x^3 + 7x^2 + 11x + 12) \div (x + 1) \)

Step2: Set Up Synthetic Division

Coefficients: \( 11, 12, 7, 11, 12 \). \( c = -1 \) (since \( x + 1 = x - (-1) \)).
Bring down \( 11 \).

Step3: Multiply and Add

• \( 11 \times (-1) = -11 \). \( 12 + (-11) = 1 \).
• \( 1 \times (-1) = -1 \). \( 7 + (-1) = 6 \).
• \( 6 \times (-1) = -6 \). \( 11 + (-6) = 5 \).
• \( 5 \times (-1) = -5 \). \( 12 + (-5) = 7 \).

Result: \( 11x^3 + x^2 + 6x + 5 + \frac{7}{x + 1} \), quotient \( 11x^3 + x^2 + 6x + 5 \), remainder \( 7 \).

Part 2a: Determine if \( d(x) = x + 1 \) is factor of \( f(x) = x^3 - 5x^2 - 7x - 2 \)

Step2: Use Factor Theorem

By Factor Theorem, \( d(x) \) is a factor if \( f(-1) = 0 \) (since \( x + 1 = x - (-1) \)).

Step3: Evaluate \( f(-1) \) via Synthetic Division

Coefficients: \( 1, -5, -7, -2 \), \( c = -1 \).
Bring down \( 1 \).

• \( 1 \times (-1) = -1 \). \( -5 + (-1) = -6 \).
• \( -6 \times (-1) = 6 \). \( -7 + 6 = -1 \).
• \( -1 \times (-1) = 1 \). \( -2 + 1 = -1 \).

Remainder is \( -1
eq 0 \), so \( d(x) \) is not a factor.

Part 2b: Determine if \( d(x) = x + 3 \) is factor of \( f(x) = 2x^4 + x^3 - 19x^2 - 17x - 15 \)

Step2: Use Factor Theorem (Synthetic Division)

\( c = -3 \) (since \( x + 3 = x - (-3) \)). Coefficients: \( 2, 1, -19, -17, -15 \).
Bring down \( 2 \).

Step3: Multiply and Add

• \( 2 \times (-3) = -6 \). \( 1 + (-6) = -5 \).
• \( -5 \times (-3) = 15 \). \( -19 + 15 = -4 \).
• \( -4 \times (-3) = 12 \). \( -17 + 12 = -5 \).
• \( -5 \times (-3) = 15 \). \( -15 + 15 = 0 \).

Remainder is \( 0 \), so \( d(x) \) is a factor.

Answer:

1a) Quotient: \( 5x^2 + 6x \), Remainder: \( -1 \)
1b) Quotient: \( 11x^3 + x^2 + 6x + 5 \), Remainder: \( 7 \)
2a) \( d(x) \) is not a factor of \( f(x) \)
2b) \( d(x) \) is a factor of \( f(x) \)