Step 1: Relate the roots p, q, r to polynomial coefficients using Vieta's Formulas

Expanding the left side of the identity gives:


\((x-p)(x-q)(x-r)=x^3-(p+q+r) x^2+(p q+q r+p r) x-p q r\)

Comparing this to the right side $x^3-16 x^2+k x-140$, we set up Vieta's formulas for positive integers p, q, r :

1. Sum of roots:


\(p+q+r=16\)

2. Product of roots:


\(p \cdot q \cdot r=140\)

3. Coefficient \(\boldsymbol{k}\) (pairwise product sum):

\(k=p q+q r+p r\)

Step 2: Find the unique positive integers p, q, r

Find the prime factorization of 140 :

\(140=2^2 \times 5 \times 7=2 \times 2 \times 5 \times 7\)


We need to group these prime factors into three positive integer factors p, q, r such that their sum equals 16:

Test combinations of three factors of 140:

If p=1 , then \(q \cdot r=140\) , but \(p+q+r=1+q+r=16 \Longrightarrow q+r=15 \) . The factors of 140 summing to 15 are not possible.
Try \( p=2: q \cdot r=70 \) Sum needed: q+r=14 .

Factors of 70: (1, 70), (2, 35), (5, 14), (7, 10). None of these pairs sum to 14.

Try \( p=4: q \cdot r=35\) . Sum needed: q+r=12 .

Factors of 35: \((5,7) \Longrightarrow 5+7=12\) . Perfect!

Thus, the positive integer roots are:

\(p=4, \quad q=5, \quad r=7\)

Check:
Sum: \(4+5+7=16\) (matches coefficient of \( x^2\) )



Product: \(4 \times 5 \times 7=140\) (matches constant term)

Step 3: Calculate the value of k

Using Vieta's formula for \(k=p q+q r+p r\) :

\(k=(4 \times 5)+(5 \times 7)+(4 \times 7)\)

\(k=20+35+28=83\)


Correct Answer
(B) 83