The polynomial $ax^{3}+bx^{2}+x-6$ has $(x-2)$ as a factor and leaves remainder 4 when divided by $(x-2)$. Find the value of a and b.

Question

Accepted Answer

The question asks us to find the values of $a$ and $b$ for the polynomial $P(x) = ax^{3}+bx^{2}+x-6$ based on two given conditions:
1. $(x-2)$ is a factor of the polynomial.
2. The polynomial leaves a remainder of 4 when divided by $(x-2)$.

Let's analyze these conditions using the Remainder Theorem and Factor Theorem.

**Understanding the Conditions:**

*   **Remainder Theorem:** When a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$.
*   **Factor Theorem:** If $(x-c)$ is a factor of a polynomial $P(x)$, then $P(c)=0$. This is a special case of the Remainder Theorem where the remainder is 0.

From the second condition, "the polynomial leaves a remainder of 4 when divided by $(x-2)$", we can apply the Remainder Theorem.
This means $P(2) = 4$.

Let's substitute $x=2$ into the polynomial $P(x) = ax^{3}+bx^{2}+x-6$:
$P(2) = a(2)^3 + b(2)^2 + (2) - 6$
$P(2) = 8a + 4b + 2 - 6$
$P(2) = 8a + 4b - 4$

Now, we equate this expression for $P(2)$ to the given remainder, which is 4:
$8a + 4b - 4 = 4$
Add 4 to both sides:
$8a + 4b = 8$
Divide the entire equation by 4 to simplify:
$2a + b = 2$

This is the equation that $a$ and $b$ must satisfy based on the remainder condition. Now, we will check each option to see which pair of $(a, b)$ values satisfies this equation.

**Checking the Options:**

**1. Option 1: $a=1, b=1$**
Substitute these values into the equation $2a+b=2$:
$2(1) + 1 = 2 + 1 = 3$
Since $3 
eq 2$, Option 1 is incorrect.

**2. Option 2: $a=0, b=2$**
Substitute these values into the equation $2a+b=2$:
$2(0) + 2 = 0 + 2 = 2$
Since $2 = 2$, Option 2 is correct.

**3. Option 3: $a=2, b=0$**
Substitute these values into the equation $2a+b=2$:
$2(2) + 0 = 4 + 0 = 4$
Since $4 
eq 2$, Option 3 is incorrect.

**4. Option 4: $a=7, b=5$**
Substitute these values into the equation $2a+b=2$:
$2(7) + 5 = 14 + 5 = 19$
Since $19 
eq 2$, Option 4 is incorrect.

Based on the condition that the polynomial leaves a remainder of 4 when divided by $(x-2)$, only Option 2 ($a=0, b=2$) satisfies the derived equation $2a+b=2$.

**Addressing the first condition: "has $(x-2)$ as a factor"**
If $(x-2)$ were a factor of the polynomial, then according to the Factor Theorem, $P(2)$ would be 0. However, our calculations based on the given remainder show $P(2)=4$. This means that for the values $a=0$ and $b=2$ (which are derived from the remainder condition), the polynomial is $P(x) = 0x^3 + 2x^2 + x - 6 = 2x^2 + x - 6$. For this polynomial, $P(2) = 2(2)^2 + 2 - 6 = 8 + 2 - 6 = 4$. Since $P(2)=4 
eq 0$, $(x-2)$ is not a factor of this polynomial. The problem statement presents two conditions, and the condition regarding the remainder of 4 is the one that leads to the unique solution provided in Option 2.

The final answer is $\boxed{	ext{Option 2}}$.