A quadratic polynomial ax 2+ bx + c = 0 is such that when it is divided by x, (x - 1) and (x + 1), the remainders are 3, 6 and 4 respectively. What is the value of (a + b)?
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Solving Quadratic Polynomial Remainder Problems
The problem asks us to find the value of (a + b) for a quadratic polynomial \(P(x) = ax^2 + bx + c\). We are given the remainders when the polynomial is divided by specific linear expressions: x, (x - 1), and (x + 1).
Understanding the Remainder Theorem
The key to solving this problem is the Remainder Theorem. This theorem states that if a polynomial \(P(x)\) is divided by a linear polynomial \((x - k)\), then the remainder of the division is equal to \(P(k)\).
Let's apply this theorem to the given information:
- When \(P(x)\) is divided by x (which is the same as \(x - 0\)), the remainder is 3. According to the Remainder Theorem, this means \(P(0) = 3\).
- When \(P(x)\) is divided by (x - 1), the remainder is 6. According to the Remainder Theorem, this means \(P(1) = 6\).
- When \(P(x)\) is divided by (x + 1) (which is the same as \(x - (-1)\)), the remainder is 4. According to the Remainder Theorem, this means \(P(-1) = 4\).
Setting Up Equations
Now, let's substitute the values into the polynomial \(P(x) = ax^2 + bx + c\):
- \(P(0) = a(0)^2 + b(0) + c = 0 + 0 + c = c\)
- \(P(1) = a(1)^2 + b(1) + c = a + b + c\)
- \(P(-1) = a(-1)^2 + b(-1) + c = a(1) - b + c = a - b + c\)
Using the given remainder values, we get the following equations:
- \(c = 3\) (from \(P(0) = 3\))
- \(a + b + c = 6\) (from \(P(1) = 6\))
- \(a - b + c = 4\) (from \(P(-1) = 4\))
Solving for a and b (and c)
From equation (1), we already know that \(c = 3\).
Substitute the value of c into equations (2) and (3):
- Equation (2) becomes: \(a + b + 3 = 6\)
- Subtract 3 from both sides: \(a + b = 6 - 3 \implies a + b = 3\)
- Equation (3) becomes: \(a - b + 3 = 4\)
- Subtract 3 from both sides: \(a - b = 4 - 3 \implies a - b = 1\)
We now have a system of two linear equations with two variables, a and b:
\(a + b = 3\) (Equation A)
\(a - b = 1\) (Equation B)
The question asks specifically for the value of \((a + b)\). Looking at Equation A, we can see directly that \(a + b = 3\).
Alternatively, we could solve for a and b individually:
Add Equation A and Equation B:
\((a + b) + (a - b) = 3 + 1\)
\(2a = 4\)
\(a = \frac{4}{2} = 2\)
Substitute the value of a (2) into Equation A:
\(2 + b = 3\)
\(b = 3 - 2 = 1\)
So, we find that \(a=2\), \(b=1\), and \(c=3\). The polynomial is \(2x^2 + x + 3\). We can verify this by checking the remainders:
- \(P(0) = 2(0)^2 + 0 + 3 = 3\) (Correct)
- \(P(1) = 2(1)^2 + 1 + 3 = 2 + 1 + 3 = 6\) (Correct)
- \(P(-1) = 2(-1)^2 + (-1) + 3 = 2 - 1 + 3 = 4\) (Correct)
The value of \((a + b)\) is \(2 + 1 = 3\).
Final Answer
Based on our calculations using the Remainder Theorem, the value of \((a + b)\) is 3.
| Condition | Remainder Theorem Application | Equation |
|---|---|---|
| Divided by x | \(P(0) = 3\) | \(c = 3\) |
| Divided by (x - 1) | \(P(1) = 6\) | \(a + b + c = 6\) |
| Divided by (x + 1) | \(P(-1) = 4\) | \(a - b + c = 4\) |
Using \(c=3\):
- \(a + b + 3 = 6 \implies a + b = 3\)
- \(a - b + 3 = 4 \implies a - b = 1\)
From \(a + b = 3\), we get the required value directly.
Revision Table: Polynomial Remainder Concepts
| Concept | Description | Relation to Problem |
|---|---|---|
| Quadratic Polynomial | A polynomial of degree 2, form \(ax^2 + bx + c\). | The specific type of polynomial given in the question. |
| Remainder Theorem | \(P(x)\) divided by \((x - k)\) has remainder \(P(k)\). | Central theorem used to derive equations from the given remainders. |
| Division by x | Equivalent to division by \((x - 0)\). Remainder is \(P(0)\). | Used to find the value of the constant term c. |
| Division by (x - 1) | Remainder is \(P(1)\). | Used to form an equation involving a, b, and c. |
| Division by (x + 1) | Equivalent to division by \((x - (-1))\). Remainder is \(P(-1)\). | Used to form another equation involving a, b, and c. |
Additional Information: Polynomial Properties
Polynomials are algebraic expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. A quadratic polynomial has the general form \(ax^2 + bx + c\), where a, b, and c are coefficients and a is not equal to 0. The degree of a polynomial is the highest power of the variable in the expression.
The Remainder Theorem is a direct consequence of the Polynomial Remainder Theorem, which states that for any polynomial \(P(x)\) and any number k, there exists a polynomial \(Q(x)\) such that \(P(x) = (x - k)Q(x) + R\), where R is the remainder. When \((x - k)\) is a linear factor, the remainder R is a constant, and substituting \(x = k\) gives \(P(k) = (k - k)Q(k) + R = 0 \cdot Q(k) + R = R\).
This problem demonstrates a common application of the Remainder Theorem to find unknown coefficients of a polynomial when information about its remainders upon division is provided.
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