`(ax+4)(bx-1)=21x^(2)+kx-4`
If the equation above is true for all values of x and a+b=10, what are the possible values for the constant k?

Options

A. 8 and 14
B. 5 and 25
C. 9 and 15
D. 19 and 31

Answer

To solve the equation \((ax + 4)(bx - 1) = 21x^2 + kx - 4\) for the constant \(k\), given that \(a + b = 10\), we will follow these steps:

### Step 1: Expand the left-hand side of the equation
We start by expanding the left-hand side:
\[ (ax + 4)(bx - 1) = abx^2 + (4b - a)x - 4 \]

### Step 2: Set the expanded form equal to the right-hand side
Now we equate the expanded left-hand side to the right-hand side:
\[ abx^2 + (4b - a)x - 4 = 21x^2 + kx - 4 \]

### Step 3: Compare coefficients
From the equation, we can compare coefficients of \(x^2\), \(x\), and the constant term:
1. Coefficient of \(x^2\): \(ab = 21\)
2. Coefficient of \(x\): \(4b - a = k\)
3. Constant term: \(-4 = -4\) (which is already satisfied)

### Step 4: Use the condition \(a + b = 10\)
We know from the problem that \(a + b = 10\). We can express \(b\) in terms of \(a\):
\[ b = 10 - a \]

### Step 5: Substitute \(b\) into the equation \(ab = 21\)
Substituting \(b\) into the equation \(ab = 21\):
\[ a(10 - a) = 21 \]
Expanding this gives:
\[ 10a - a^2 = 21 \]
Rearranging the equation:
\[ -a^2 + 10a - 21 = 0 \]
Multiplying through by -1 to make it standard:
\[ a^2 - 10a + 21 = 0 \]

### Step 6: Factor the quadratic equation
Next, we factor the quadratic:
\[ (a - 3)(a - 7) = 0 \]
This gives us two possible values for \(a\):
\[ a = 3 \quad \text{or} \quad a = 7 \]

### Step 7: Find corresponding values of \(b\)
Using \(b = 10 - a\):
- If \(a = 3\), then \(b = 10 - 3 = 7\).
- If \(a = 7\), then \(b = 10 - 7 = 3\).

### Step 8: Calculate \(k\) for both pairs \((a, b)\)
Now we can find \(k\) using \(4b - a = k\):
1. For \(a = 3\) and \(b = 7\):
\[ k = 4(7) - 3 = 28 - 3 = 25 \]
2. For \(a = 7\) and \(b = 3\):
\[ k = 4(3) - 7 = 12 - 7 = 5 \]

### Conclusion
The possible values for the constant \(k\) are:
\[ \boxed{5 \text{ and } 25} \]