Write the function in the form \( f(x)=(x-k) q(x)+r(x) \) for the given value of \( k \). \( f(x)=x^{3}+5 x^{2}-5 x-17, k=-\sqrt{5} \) \( f(x)=\square \) Demonstrate that \( f(k)=r \). \( f(-\sqrt{5})= \)

We are given   f(x) = x³ + 5x² – 5x – 17 and the value   k = –√5. We want to write f(x) in the form   f(x) = (x – k)·q(x) + r(x). Since k = –√5, note that x – k = x – (–√5) = x + √5. Also, when dividing by a linear factor, the remainder r(x) will be a constant r. So we write   f(x) = (x + √5)·q(x) + r. The Remainder Theorem tells us that r = f(–√5). Let’s compute that: 1. Substitute x = –√5 into f(x):   f(–√5) = (–√5)³ + 5(–√5)² – 5(–√5) – 17. 2. Calculate each term:   (–√5)³ = –(√5)³ = –(5√5)     (since (√5)³ = 5√5)   5(–√5)² = 5·(5) = 25           (since (–√5)² = 5)   –5(–√5) = 5√5,   –17 remains as –17. 3. Now, add these together:   f(–√5) = (–5√5) + 25 + (5√5) – 17.   The –5√5 and +5√5 cancel, leaving:   f(–√5) = 25 – 17 = 8. Thus, the remainder is r = 8. Next, we need to determine the quotient polynomial q(x), which must be quadratic (since f(x) is cubic). Let   q(x) = Ax² + Bx + C. Then our representation becomes:   f(x) = (x + √5)(Ax² + Bx + C) + 8. Expand the product:   (x + √5)(Ax² + Bx + C) = Ax³ + Bx² + Cx + A√5·x² + B√5·x + C√5 Combine like terms to get:   = Ax³ + (B + A√5)x² + (C + B√5)x + C√5. So, we have:   f(x) = Ax³ + (B + A√5)x² + (C + B√5)x + (C√5 + 8). This must equal the original f(x) = x³ + 5x² – 5x – 17. Equate the coefficients term-by-term: 1. Coefficient of x³:   A = 1. 2. Coefficient of x²:   B + A√5 = B + √5 = 5  ⟹ B = 5 – √5. 3. Coefficient of x:   C + B√5 = C + (5 – √5)√5 = C + 5√5 – 5 = –5. Solve for C:   C = –5 – 5√5 + 5 = –5√5. 4. Constant term:   C√5 + 8 = (–5√5)√5 + 8 = –5·5 + 8 = –25 + 8 = –17. This checks out with the constant term of f(x). Thus, the quotient polynomial is:   q(x) = x² + (5 – √5)x – 5√5. Finally, we can write the final decomposition of f(x) as:   f(x) = (x + √5)·[x² + (5 – √5)x – 5√5] + 8. And as demonstrated earlier, f(–√5) = 8. So, the answer is:   f(x) = (x + √5)(x² + (5 – √5)x – 5√5) + 8,   with f(–√5) = 8.