Consider the equations `(x-p)(x-6)+1 = 0` having integral coefficents. If the equation has integral roots, then what values can p have?

Options

A. 4 or 8
B. 5 or 10
C. 6 or 12
D. 3 or 6

Explanation

To solve the equation \((x - p)(x - 6) + 1 = 0\) for integral values of \(p\) such that the equation has integral roots, we can follow these steps:

### Step 1: Expand the equation
We start by expanding the given equation:
\[ (x - p)(x - 6) + 1 = 0 \]
Expanding the product:
\[ x^2 - 6x - px + 6p + 1 = 0 \]
This simplifies to:
\[ x^2 - (p + 6)x + (6p + 1) = 0 \]

### Step 2: Identify coefficients
From the expanded form, we can identify the coefficients:
- \(a = 1\)
- \(b = -(p + 6)\)
- \(c = 6p + 1\)

### Step 3: Use the condition for integral roots
For the quadratic equation to have integral roots, the discriminant must be a perfect square. The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (-(p + 6))^2 - 4 \cdot 1 \cdot (6p + 1) \]
This simplifies to:
\[ D = (p + 6)^2 - 4(6p + 1) \]
Expanding this:
\[ D = p^2 + 12p + 36 - (24p + 4) \]
Combining like terms:
\[ D = p^2 - 12p + 32 \]

### Step 4: Set the discriminant as a perfect square
For \(D\) to be a perfect square, we set:
\[ p^2 - 12p + 32 = k^2 \]
where \(k\) is an integer.

### Step 5: Rearranging the equation
Rearranging gives us:
\[ p^2 - 12p + (32 - k^2) = 0 \]
This is a quadratic equation in \(p\). The discriminant of this quadratic must also be a perfect square:
\[ D' = (-12)^2 - 4 \cdot 1 \cdot (32 - k^2) = 144 - 128 + 4k^2 = 16 + 4k^2 \]
Setting \(D'\) as a perfect square:
\[ 16 + 4k^2 = m^2 \]
for some integer \(m\).

### Step 6: Solving the equation
Rearranging gives:
\[ m^2 - 4k^2 = 16 \]
This can be factored as:
\[ (m - 2k)(m + 2k) = 16 \]
The pairs of factors of 16 are:
- (1, 16)
- (2, 8)
- (4, 4)
- (-1, -16)
- (-2, -8)
- (-4, -4)

### Step 7: Finding integer values of \(p\)
For each factor pair, we can solve for \(m\) and \(k\) and then substitute back to find possible values of \(p\).

After calculating for each pair, we find that the possible integer values of \(p\) that satisfy the original equation with integral roots are:
\[ p = 4, 8 \]

### Final Result
The values of \(p\) that allow the equation \((x - p)(x - 6) + 1 = 0\) to have integral roots are \(p = 4\) and \(p = 8\).