If `(ax+2)(bx+7)=15x^(2)+cx+14` for all values of x, and `a+b=8`, what are the two possible values for c ?

Options

A. 3 and 5
B. 6 and 35
C. 10 and 21
D. 31 and 41

Answer

To solve the equation \((ax + 2)(bx + 7) = 15x^2 + cx + 14\) for all values of \(x\) and given that \(a + b = 8\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:
\[ (ax + 2)(bx + 7) = abx^2 + 7ax + 2bx + 14 \]
This simplifies to:
\[ abx^2 + (7a + 2b)x + 14 \]

### Step 2: Set up the equation
Now we can set the expanded left-hand side equal to the right-hand side:
\[ abx^2 + (7a + 2b)x + 14 = 15x^2 + cx + 14 \]

### Step 3: Compare coefficients
By comparing the coefficients of \(x^2\), \(x\), and the constant terms, we get the following equations:
1. \(ab = 15\) (from the coefficient of \(x^2\))
2. \(7a + 2b = c\) (from the coefficient of \(x\))
3. The constant terms are equal, so they do not provide new information.

### Step 4: Use the given condition \(a + b = 8\)
From the condition \(a + b = 8\), we can express \(a\) in terms of \(b\):
\[ a = 8 - b \]

### Step 5: Substitute \(a\) in the first equation
Substituting \(a\) into the first equation \(ab = 15\):
\[ (8 - b)b = 15 \]
This simplifies to:
\[ 8b - b^2 = 15 \]
Rearranging gives:
\[ b^2 - 8b + 15 = 0 \]

### Step 6: Factor the quadratic equation
Now we can factor the quadratic equation:
\[ (b - 5)(b - 3) = 0 \]
This gives us two possible values for \(b\):
\[ b = 5 \quad \text{or} \quad b = 3 \]

### Step 7: Find corresponding values of \(a\)
Using \(a + b = 8\):
- If \(b = 5\), then \(a = 8 - 5 = 3\).
- If \(b = 3\), then \(a = 8 - 3 = 5\).

### Step 8: Calculate \(c\) for both pairs \((a, b)\)
Now we can calculate \(c\) using \(7a + 2b\):
1. For \(a = 3\) and \(b = 5\):
\[ c = 7(3) + 2(5) = 21 + 10 = 31 \]
2. For \(a = 5\) and \(b = 3\):
\[ c = 7(5) + 2(3) = 35 + 6 = 41 \]

### Conclusion
The two possible values for \(c\) are:
\[ \boxed{31} \quad \text{and} \quad \boxed{41} \]