159k views
0 votes
When dividing the polynomial (x³ – 2x²+4 )/( x +1) using synthetic division, which of the following setup boxes would be used?

User Novellino
by
7.6k points

1 Answer

2 votes

The correct setup box for the synthetic division is:

-1 | 1 -2 4

| 1 (coefficient of x^3) (coefficient of x^2) (coefficient of x) (constant term)

-(x +1)

| 1 (-1 * 1) (-1 * (-1) + 1) (-1 * 4 + 1)

------

| 1 -3 3 -3

Step 1: Set up the divisor and dividend coefficients.

| Divisor: x + 1 |

|---|---|

| Constant term: 1 |

| Dividend: x³ – 2x² + 4 |

|---|---|

| x³ coefficient: 1 |

| x² coefficient: -2 |

| x coefficient: 4 |

| Constant term: 4 |

Step 2: Place the divisor coefficient (negative coefficient of x) as the first term inside the table.

| -1: | 1 | -2 | 4 |

Step 3: Bring down the x³ coefficient of the dividend.

| -1: | 1 | -2 | 4 |

| ↓ | 1 | | |

Step 4: Multiply the top term by the brought-down coefficient and place the product below the next coefficient of the dividend (x²).

| -1: | 1 | -2 | 4 |

| ↓ | 1 | -1 | |

Step 5: Add the product and the next coefficient of the dividend. Write the sum below.

| -1: | 1 | -3 | 4 |

| ↓ | 1 | | |

Step 6: Repeat steps 4 and 5 for the remaining coefficients of the dividend.

| -1: | 1 | -3 | 3 |

| ↓ | 1 | | -3 |

Step 7: The bottom row represents the coefficients of the quotient polynomial in reverse order.

| -1: | 1 | -3 | 3 |

| ↓ | | 1 | -3 | -3 |

Therefore, the correct setup box for the synthetic division is:

-1 | 1 -2 4

| 1 (coefficient of x^3) (coefficient of x^2) (coefficient of x) (constant term)

-(x +1)

| 1 (-1 * 1) (-1 * (-1) + 1) (-1 * 4 + 1)

------

| 1 -3 3 -3

User Nantoka
by
7.9k points