Here are the step-by-step solutions for each problem using synthetic division.
1) $(r^3 + 6r^2 - 21r - 18) \div (r - 3)$
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Setup: The divisor is $r - 3$, so we put positive $3$ in the box. The coefficients are $1, 6, -21, -18$.
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Step 1: Bring down the $1$.
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Step 2: Multiply $1 \times 3 = 3$. Add to next term: $6 + 3 = 9$.
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Step 3: Multiply $9 \times 3 = 27$. Add to next term: $-21 + 27 = 6$.
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Step 4: Multiply $6 \times 3 = 18$. Add to next term: $-18 + 18 = 0$.
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Result: The numbers at the bottom are $1, 9, 6$ with a remainder of $0$. Since we started with $r^3$, the answer starts with $r^2$.
2) $(x^3 - 11x^2 + 22x + 40) \div (x - 5)$
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Setup: The divisor is $x - 5$, so we use positive $5$. Coefficients: $1, -11, 22, 40$.
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Step 1: Bring down the $1$.
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Step 2: Multiply $1 \times 5 = 5$. Add: $-11 + 5 = -6$.
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Step 3: Multiply $-6 \times 5 = -30$. Add: $22 + (-30) = -8$.
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Step 4: Multiply $-8 \times 5 = -40$. Add: $40 + (-40) = 0$.
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Result: Coefficients are $1, -6, -8$. Remainder is $0$.
3) $(9x^3 - 19x^2 - 28x + 12) \div (x - 3)$
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Setup: Divisor is $x - 3$, so use $3$. Coefficients: $9, -19, -28, 12$.
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Step 1: Bring down the $9$.
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Step 2: Multiply $9 \times 3 = 27$. Add: $-19 + 27 = 8$.
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Step 3: Multiply $8 \times 3 = 24$. Add: $-28 + 24 = -4$.
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Step 4: Multiply $-4 \times 3 = -12$. Add: $12 + (-12) = 0$.
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Result: Coefficients are $9, 8, -4$. Remainder is $0$.
4) $(m^3 - 13m^2 + 24m + 18) \div (m - 3)$
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Setup: Divisor is $m - 3$, so use $3$. Coefficients: $1, -13, 24, 18$.
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Step 1: Bring down the $1$.
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Step 2: Multiply $1 \times 3 = 3$. Add: $-13 + 3 = -10$.
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Step 3: Multiply $-10 \times 3 = -30$. Add: $24 + (-30) = -6$.
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Step 4: Multiply $-6 \times 3 = -18$. Add: $18 + (-18) = 0$.
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Result: Coefficients are $1, -10, -6$. Remainder is $0$.
5) $(x^3 + 15x^2 + 45x - 25) \div (x + 5)$
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Setup: The divisor is $x + 5$, so we use negative $-5$. Coefficients: $1, 15, 45, -25$.
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Step 1: Bring down the $1$.
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Step 2: Multiply $1 \times -5 = -5$. Add: $15 + (-5) = 10$.
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Step 3: Multiply $10 \times -5 = -50$. Add: $45 + (-50) = -5$.
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Step 4: Multiply $-5 \times -5 = 25$. Add: $-25 + 25 = 0$.
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Result: Coefficients are $1, 10, -5$. Remainder is $0$.
6) $(a^3 + 5a^2 + 14a + 16) \div (a + 2)$
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Setup: The divisor is $a + 2$, so we use negative $-2$. Coefficients: $1, 5, 14, 16$.
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Step 1: Bring down the $1$.
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Step 2: Multiply $1 \times -2 = -2$. Add: $5 + (-2) = 3$.
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Step 3: Multiply $3 \times -2 = -6$. Add: $14 + (-6) = 8$.
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Step 4: Multiply $8 \times -2 = -16$. Add: $16 + (-16) = 0$.
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Result: Coefficients are $1, 3, 8$. Remainder is $0$.
Final Answer:
1) $r^2 + 9r + 6$
2) $x^2 - 6x - 8$
3) $9x^2 + 8x - 4$
4) $m^2 - 10m - 6$
5) $x^2 + 10x - 5$
6) $a^2 + 3a + 8$
Parent Tip: Review the logic above to help your child master the concept of dividing polynomials using synthetic division worksheet answer key.
