 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/3/solution

f (1) = 23 - 2 p

. Equating it to

9

gives $p = 7$ .

For the separate polynomial long division, successively cancel the leading powers of the dividend. Subtracting

x^{2} (x - 1)

2 x^{2} + 2 x + 1

; subtracting

2 x (x - 1)

4 x + 1

; subtracting

4 (x - 1)

5

. Thus

x^{3} + x^{2} + 2 x + 1 = (x - 1) (x^{2} + 2 x + 4) + 5, quotient x^{2} + 2 x + 4, remainder 5.

The polynomial remainder also equals the dividend evaluated at

x = 1

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