Solved Dividing Polynomials Sheet1 Divide by synthetic | Chegg.com - Free Printable

Here are the step-by-step solutions for each problem using synthetic division.

1) $(n^2 + 6n - 27) \div (n - 3)$

* Setup: The divisor is $n - 3$, so we use $3$ on the outside. The coefficients of the dividend are $1, 6, -27$.
* Step 1: Bring down the $1$.
* Step 2: Multiply $1 \times 3 = 3$. Add to the next coefficient: $6 + 3 = 9$.
* Step 3: Multiply $9 \times 3 = 27$. Add to the next coefficient: $-27 + 27 = 0$.
* Result: The remainder is $0$. The quotient coefficients are $1, 9$. This corresponds to $1n + 9$.

2) $(2b^4 + 7b^3 + 12b^2 - 9b - 27) \div (2b + 3)$

* Setup: The divisor is $2b + 3$. To use synthetic division easily, we divide by $(b + \frac{3}{2})$, so we use $-\frac{3}{2}$ (or $-1.5$) on the outside. Coefficients: $2, 7, 12, -9, -27$.
* Step 1: Bring down $2$.
* Step 2: $2 \times (-1.5) = -3$. Add to $7$: $7 + (-3) = 4$.
* Step 3: $4 \times (-1.5) = -6$. Add to $12$: $12 + (-6) = 6$.
* Step 4: $6 \times (-1.5) = -9$. Add to $-9$: $-9 + (-9) = -18$.
* Step 5: $-18 \times (-1.5) = 27$. Add to $-27$: $-27 + 27 = 0$.
* Adjustment: The numbers at the bottom are $2, 4, 6, -18$. These are coefficients for dividing by $(b + 1.5)$. Since we originally divided by $(2b + 3)$, which is $2(b + 1.5)$, we must divide these resulting coefficients by $2$ to get the true quotient.
* $2 \div 2 = 1$
* $4 \div 2 = 2$
* $6 \div 2 = 3$
* $-18 \div 2 = -9$
* Result: The quotient is $b^3 + 2b^2 + 3b - 9$.

3) $(2t^4 + t^3 + 5t^2 + 14t - 16) \div (t + 2)$

* Setup: Divisor is $t + 2$, so use $-2$. Coefficients: $2, 1, 5, 14, -16$.
* Step 1: Bring down $2$.
* Step 2: $2 \times (-2) = -4$. Add to $1$: $1 + (-4) = -3$.
* Step 3: $-3 \times (-2) = 6$. Add to $5$: $5 + 6 = 11$.
* Step 4: $11 \times (-2) = -22$. Add to $14$: $14 + (-22) = -8$.
* Step 5: $-8 \times (-2) = 16$. Add to $-16$: $-16 + 16 = 0$.
* Result: Coefficients are $2, -3, 11, -8$. Remainder is $0$.
* Answer: $2t^3 - 3t^2 + 11t - 8$.

4) $(4v^3 + 11v^2 - 25) \div (4v - 5)$

* Setup: Note that the $v$ term is missing in the dividend, so we must use a $0$ coefficient. Dividend coefficients: $4, 11, 0, -25$. Divisor is $4v - 5$. We divide by $(v - \frac{5}{4})$, so use $1.25$ on the outside.
* Step 1: Bring down $4$.
* Step 2: $4 \times 1.25 = 5$. Add to $11$: $11 + 5 = 16$.
* Step 3: $16 \times 1.25 = 20$. Add to $0$: $0 + 20 = 20$.
* Step 4: $20 \times 1.25 = 25$. Add to $-25$: $-25 + 25 = 0$.
* Adjustment: Bottom row is $4, 16, 20$. Since we divided by $(v - 1.25)$ but the original divisor was $(4v - 5)$, we divide the resulting coefficients by $4$.
* $4 \div 4 = 1$
* $16 \div 4 = 4$
* $20 \div 4 = 5$
* Result: $v^2 + 4v + 5$.

5) $(h^3 - 6h^2 + 32) \div (h - 4)$

* Setup: The $h$ term is missing, so use coefficient $0$. Coefficients: $1, -6, 0, 32$. Divisor is $h - 4$, so use $4$.
* Step 1: Bring down $1$.
* Step 2: $1 \times 4 = 4$. Add to $-6$: $-6 + 4 = -2$.
* Step 3: $-2 \times 4 = -8$. Add to $0$: $0 + (-8) = -8$.
* Step 4: $-8 \times 4 = -32$. Add to $32$: $32 + (-32) = 0$.
* Result: Coefficients are $1, -2, -8$. Remainder is $0$.
* Answer: $h^2 - 2h - 8$.

6) $(q^2 + 10q + 24) \div (q + 6)$

* Setup: Divisor is $q + 6$, so use $-6$. Coefficients: $1, 10, 24$.
* Step 1: Bring down $1$.
* Step 2: $1 \times (-6) = -6$. Add to $10$: $10 + (-6) = 4$.
* Step 3: $4 \times (-6) = -24$. Add to $24$: $24 + (-24) = 0$.
* Result: Coefficients are $1, 4$. Remainder is $0$.
* Answer: $q + 4$.

7) $(d^3 + 13d^2 + 40d + 36) \div (d + 9)$

* Setup: Divisor is $d + 9$, so use $-9$. Coefficients: $1, 13, 40, 36$.
* Step 1: Bring down $1$.
* Step 2: $1 \times (-9) = -9$. Add to $13$: $13 + (-9) = 4$.
* Step 3: $4 \times (-9) = -36$. Add to $40$: $40 + (-36) = 4$.
* Step 4: $4 \times (-9) = -36$. Add to $36$: $36 + (-36) = 0$.
* Result: Coefficients are $1, 4, 4$. Remainder is $0$.
* Answer: $d^2 + 4d + 4$.

8) $(p^4 - 6p^3 - 15p^2 - 12p + 32) \div (p - 8)$

* Setup: Divisor is $p - 8$, so use $8$. Coefficients: $1, -6, -15, -12, 32$.
* Step 1: Bring down $1$.
* Step 2: $1 \times 8 = 8$. Add to $-6$: $-6 + 8 = 2$.
* Step 3: $2 \times 8 = 16$. Add to $-15$: $-15 + 16 = 1$.
* Step 4: $1 \times 8 = 8$. Add to $-12$: $-12 + 8 = -4$.
* Step 5: $-4 \times 8 = -32$. Add to $32$: $32 + (-32) = 0$.
* Result: Coefficients are $1, 2, 1, -4$. Remainder is $0$.
* Answer: $p^3 + 2p^2 + p - 4$.

Final Answer:
1) $n + 9$
2) $b^3 + 2b^2 + 3b - 9$
3) $2t^3 - 3t^2 + 11t - 8$
4) $v^2 + 4v + 5$
5) $h^2 - 2h - 8$
6) $q + 4$
7) $d^2 + 4d + 4$
8) $p^3 + 2p^2 + p - 4$