Polynomial Factors and Coefficients Calculation

The problem requires finding the values of coefficients $a$ and $b$ for the polynomial $P(x) = ax^3 + 3x^2 + bx$, given that $(x + 1)$ and $(x + 2)$ are its factors.

Applying the Factor Theorem

The Factor Theorem states that if $(x - c)$ is a factor of a polynomial $P(x)$, then $P(c) = 0$. In this case, the factors are $(x + 1)$ and $(x + 2)$, which means $c = -1$ and $c = -2$ are the roots of the polynomial.

• For the factor $(x + 1)$, we have $x = -1$. Thus, $P(-1) = 0$.
• For the factor $(x + 2)$, we have $x = -2$. Thus, $P(-2) = 0$.

Formulating Equations

Substitute the roots into the polynomial equation $P(x) = ax^3 + 3x^2 + bx$:

1. When $x = -1$: $P(-1) = a(-1)^3 + 3(-1)^2 + b(-1)$ $0 = -a + 3 - b$ This simplifies to the equation: $a + b = 3$ (Equation 1)
2. When $x = -2$: $P(-2) = a(-2)^3 + 3(-2)^2 + b(-2)$ $0 = -8a + 12 - 2b$ Dividing by 2, we get: $-4a + 6 - b = 0$ This simplifies to the equation: $4a + b = 6$ (Equation 2)

Solving for Coefficients $a$ and $b$

We now have a system of two linear equations with two variables:

1. $a + b = 3$

2. $4a + b = 6$

Subtract Equation 1 from Equation 2 to eliminate $b$:

$(4a + b) - (a + b) = 6 - 3$

$3a = 3$

$a = \frac{3}{3}$

$a = 1$

Substitute the value of $a = 1$ back into Equation 1:

$1 + b = 3$

$b = 3 - 1$

$b = 2$

Conclusion

The values obtained are $a = 1$ and $b = 2$. This matches the first option.