# math problem 5?

The remainder when the polynomial P(x) is divided by x−1 and x+1 are is 4 and 8, respectively. If the remainder when P(x) is divided by x^2−1 is R(x), determine the value of R(6).

### 2 Answers

- Φ² = Φ+1Lv 74 weeks ago
x^2−1 is quadratic so R(x) is linear.

R(x) passes through (1,4) and (-1,8) so R(x) = -2x + 6, and R(6) = -6

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- Ian HLv 71 month ago
The remainders when the polynomial P(x) is divided by x − 1 and x + 1

are 4 and 8, respectively. If the remainder when P(x) is divided by x^2 − 1 is R(x), determine the value of R(6)

Suppose P = x^2 – 1 + kx + c

When the polynomial P(x) is divided by the linear divisor (x - 1) the remainder,

is equal to the value of the polynomial evaluated at x = 1.

1 – 1 + k + c = c + k = 4

When the polynomial P(x) is divided by the linear divisor (x + 1) the remainder,

is equal to the value of the polynomial evaluated at x = -1.

1 – 1 - k + c = c – k = 8

c = 6, k = -2

P = x^2 – 2x + 5

P = x^2 – 1 + (6 – 2x)

P(x) divided by x^2−1 = 1 + (6 – 2x)/(x^2 – 1)

R(x) = 6 – 2x

R(6) = -6

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