The problem you've shared asks to find a number $m$, such that the quadratic polynomial $P(x) = ax^2 + bx + c$ is divisible by $(x - m)$, given the following conditions:

1. $P(x)$ is divisible by $(x + 1)$,
2. $P(x)$ has a remainder of -3 when divided by $(x - 2)$,
3. $P(x)$ has a remainder of 4 when divided by $(x - 3)$.

Step-by-Step Solution:

Given that $P(x)$ is divisible by $(x + 1)$, we know: $P(-1) = 0$

The next condition is that when divided by $(x - 2)$, $P(x)$ leaves a remainder of -3, which means: $P(2) = -3$

Finally, the condition that $P(x)$ leaves a remainder of 4 when divided by $(x - 3)$ gives: $P(3) = 4$

Now, let's use these three conditions to set up a system of equations. Since $P(x) = ax^2 + bx + c$, we can substitute the values of $x$ from the conditions into this general form:

1. $P(-1) = 0$ $a(-1)^2 + b(-1) + c = 0 \implies a - b + c = 0$

2. $P(2) = -3$ $a(2)^2 + b(2) + c = -3 \implies 4a + 2b + c = -3$

3. $P(3) = 4$ $a(3)^2 + b(3) + c = 4 \implies 9a + 3b + c = 4$

We now have the following system of equations:

$1. a - b + c = 0$ $2. 4a + 2b + c = -3$ $3. 9a + 3b + c = 4$

Let's solve this system to find the values of $a$, $b$, and $c$.

Solving the system:

1. From equation $a - b + c = 0$, we get: $c = b - a$

2. Substitute $c = b - a$ into the other two equations:

• In $4a + 2b + c = -3$: $4a + 2b + (b - a) = -3 \implies 3a + 3b = -3 \implies a + b = -1$

• In $9a + 3b + c = 4$: $9a + 3b + (b - a) = 4 \implies 8a + 4b = 4 \implies 2a + b = 1$

Now we solve the two linear equations:

1. $a + b = -1$
2. $2a + b = 1$

Subtract equation 1 from equation 2: $(2a + b) - (a + b) = 1 - (-1) \implies a = 2$

Substitute $a = 2$ into $a + b = -1$: $2 + b = -1 \implies b = -3$

Now, substitute $a = 2$ and $b = -3$ into $c = b - a$: $c = -3 - 2 = -5$

Thus, the polynomial is: $P(x) = 2x^2 - 3x - 5$

Finding $m$:

Since $P(x)$ must be divisible by $(x - m)$, this implies $P(m) = 0$. Let's solve $P(x) = 0$ to find $m$:

$2x^2 - 3x - 5 = 0$

Using the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}$

The solutions are: $x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5$ $x = \frac{3 - 7}{4} = \frac{-4}{4} = -1$

Since $m > 0$, we have: $m = 2.5$

Thus, $m = 2.5$.

Do you want more details on any part of the process?

Here are 5 related questions to consider:

1. What are the roots of a quadratic polynomial, and how are they related to divisibility?
2. How does the remainder theorem apply to polynomials?
3. Can you explain why we use the quadratic formula in this case?
4. What would happen if $m$ were negative in the context of this problem?
5. How does solving a system of linear equations help in finding the coefficients of a polynomial?

Tip: Always check whether all the conditions of a problem are met after solving, as this helps verify the correctness of your solution.