Polynomial Definition and Given Conditions

We are working with a quadratic polynomial defined by the expression \(f(x) = ax^2 + bx + c\). The coefficients \(a\), \(b\), and \(c\) are constants.

The problem provides us with specific information about this polynomial:

• The value of the polynomial at \(x=1\) is \(2\), meaning \(f(1) = 2\).
• The value of the polynomial at \(x=4\) is also \(2\), meaning \(f(4) = 2\).
• The number \(2\) is a root of the equation \(f(x) = 0\). This implies that when we substitute \(x=2\) into the polynomial, the result is \(0\), so \(f(2) = 0\).

Calculating \((a + b + c)\) Directly

The question asks us to find the value of the specific expression \((a + b + c)\).

Let's look closely at the first condition provided: \(f(1) = 2\). The definition of the function \(f(x)\) tells us how to calculate its value for any input \(x\). To find \(f(1)\), we substitute \(x=1\) into the polynomial formula:

\(f(1) = a(1)^2 + b(1) + c\)

By performing the substitution, we get:

\(f(1) = a \times 1 + b \times 1 + c\)

\(f(1) = a + b + c\)

Now, we compare this result with the given condition \(f(1) = 2\). By equating the two expressions for \(f(1)\), we find:

\(a + b + c = 2\)

This calculation shows that the value of the expression \((a + b + c)\) is directly given by the value of the function at \(x=1\).

Verifying Consistency with All Conditions

To ensure that a polynomial satisfying all given conditions exists and to fully understand the problem, we can set up a system of equations using all the provided information and solve for the coefficients \(a\), \(b\), and \(c\).

The conditions translate into the following equations:

1. From \(f(1) = 2\): \(a(1)^2 + b(1) + c = 2 \implies a + b + c = 2\)
2. From \(f(4) = 2\): \(a(4)^2 + b(4) + c = 2 \implies 16a + 4b + c = 2\)
3. From \(f(2) = 0\): \(a(2)^2 + b(2) + c = 0 \implies 4a + 2b + c = 0\)

We can solve this system of three linear equations for \(a\), \(b\), and \(c\):

1. First, let's use equations (1) and (2). Subtract equation (1) from equation (2): \((16a + 4b + c) - (a + b + c) = 2 - 2\) \(15a + 3b = 0\) Dividing by \(3\), we get \(5a + b = 0\), which means \(b = -5a\).
2. Next, let's use equations (1) and (3). Substitute \(b = -5a\) into equation (1): \(a + (-5a) + c = 2\) \(-4a + c = 2\) We can express \(c\) in terms of \(a\): \(c = 4a + 2\).
3. Now, substitute \(b = -5a\) into equation (3): \(4a + 2(-5a) + c = 0\) \(4a - 10a + c = 0\) \(-6a + c = 0\) We can express \(c\) in terms of \(a\): \(c = 6a\).
4. We now have two expressions for \(c\). Let's set them equal to each other to find \(a\): \(6a = 4a + 2\) \(6a - 4a = 2\) \(2a = 2\) \(a = 1\)
5. With \(a = 1\), we can find \(b\) and \(c\): \(b = -5a = -5(1) = -5\) \(c = 6a = 6(1) = 6\)

The specific quadratic polynomial is therefore \(f(x) = x^2 - 5x + 6\).

Let's calculate \((a + b + c)\) using these determined coefficient values:

\(a + b + c = 1 + (-5) + 6 = 2\)

This confirms the result obtained simply by evaluating \(f(1)\).

Final Conclusion on \((a + b + c)\)

The value of the expression \((a + b + c)\) is directly equivalent to \(f(1)\). Since the problem states \(f(1) = 2\), the value of \((a + b + c)\) must be \(2\). The other conditions confirm the existence and uniqueness of such a polynomial but are not needed for this specific calculation.