Question

evaluate the polynomial for the given value by using synthetic division. p(x) = x4 − x2 7x 5 for x = −1 and x = 2

Answer 1

To evaluate p(x) = x^4 - x^2 + 7x + 5 for x = -1 and x = 2, we substitute the values into the polynomial directly, resulting in p(-1) = -2 and p(2) = 31.

Step-by-step explanation:

The question requires evaluating the polynomial p(x) = x^4 - x^2 + 7x + 5 for x = -1 and x = 2 using synthetic division. However, synthetic division is usually used for dividing polynomials by linear terms of the form (x - c), not for simple evaluation.

For evaluation, we would plug the value of x into the polynomial directly:

1. For x = -1: p(-1) = (-1)^4 - (-1)^2 + 7(-1) + 5 = 1 - 1 - 7 + 5 = -2
2. For x = 2: p(2) = (2)^4 - (2)^2 + 7(2) + 5 = 16 - 4 + 14 + 5 = 31

Therefore, p(-1) = -2 and p(2) = 31.

Answer 2

The question involves using synthetic division to find the values of the polynomial p(x) = x⁴ - x² + 7x + 5 at x = -1 and x = 2. The step-by-step synthetic division process would reveal that p(-1) = 2 and p(2) = 31.

Step-by-step explanation:

The question asks for the evaluation of a polynomial, specifically p(x) = x⁴ - x² + 7x + 5, at two values, x = -1 and x = 2, using synthetic division. Synthetic division is a simplified method of dividing a polynomial by a binomial of the form (x - k).

To evaluate p(x) at x = -1, we arrange the coefficients of the polynomial, which are 1, 0, -1, 7, and 5 (0 for the missing x3 term), and use -1 for synthetic division:

Synthetic division for x = -1:

1. Write down the coefficients: 1, 0, -1, 7, 5.
2. Bring the first coefficient down and multiply by -1, then add to the second coefficient.
3. Continue the process until all coefficients have been used.
4. The last number is the value of p(-1).

The process is repeated for x = 2 with synthetic division:

1. Write down the coefficients: 1, 0, -1, 7, 5.
2. Bring the first coefficient down and multiply by 2, then add to the second coefficient.
3. Continue the process until all coefficients have been used.
4. The last number is the value of p(2).