If highest common factor of `x^2+px-q` and `5x^2-3px-15q` is `(x-3)`, then value of `p` and `q` will be

Options

A. `-5/3,4`
B. `-5/3,-4`
C. `5/3,4`
D. `4,-3/5`

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) given that the highest common factor (HCF) of the two polynomials \( x^2 + px - q \) and \( 5x^2 - 3px - 15q \) is \( (x - 3) \).

### Step 1: Substitute \( x = 3 \) in both polynomials
Since \( (x - 3) \) is a factor, substituting \( x = 3 \) into both polynomials should yield zero.

1. For the first polynomial:
\[ 3^2 + p(3) - q = 0 \]
Simplifying this gives:
\[ 9 + 3p - q = 0 \quad \text{(Equation 1)} \]

2. For the second polynomial:
\[ 5(3^2) - 3p(3) - 15q = 0 \]
Simplifying this gives:
\[ 5(9) - 9p - 15q = 0 \]
\[ 45 - 9p - 15q = 0 \quad \text{(Equation 2)} \]

### Step 2: Rearranging the equations
From Equation 1, we can express \( q \) in terms of \( p \):
\[ q = 9 + 3p \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 2
Now, substitute \( q \) from Equation 3 into Equation 2:
\[ 45 - 9p - 15(9 + 3p) = 0 \]
Expanding this gives:
\[ 45 - 9p - 135 - 45p = 0 \]
Combining like terms results in:
\[ -54p - 90 = 0 \]

### Step 4: Solve for \( p \)
Rearranging gives:
\[ -54p = 90 \]
\[ p = -\frac{90}{54} = -\frac{5}{3} \]

### Step 5: Substitute \( p \) back to find \( q \)
Now, substitute \( p = -\frac{5}{3} \) back into Equation 3 to find \( q \):
\[ q = 9 + 3\left(-\frac{5}{3}\right) \]
This simplifies to:
\[ q = 9 - 5 = 4 \]

### Final Values
Thus, the values of \( p \) and \( q \) are:
\[ p = -\frac{5}{3}, \quad q = 4 \]