Finding Polynomial Coefficients using Factors

The problem states that the polynomial is given by $p(x) = x^3 - ax^2 + bx + 6$. We are told that $(x + 1)$ and $(x - 2)$ are factors of this polynomial.

According to the Factor Theorem, if $(x - k)$ is a factor of a polynomial $p(x)$, then $p(k) = 0$. We can use this theorem to set up equations involving the coefficients $a$ and $b$.

Applying the Factor Theorem

Since $(x + 1)$ is a factor, we know that $p(-1)$ must be equal to 0. Let's substitute $x = -1$ into the polynomial:

• $p(-1) = (-1)^3 - a(-1)^2 + b(-1) + 6$
• $0 = -1 - a(1) - b + 6$
• $0 = -1 - a - b + 6$
• $0 = 5 - a - b$
• This gives us our first equation: $a + b = 5$ (Equation 1)

Similarly, since $(x - 2)$ is a factor, we know that $p(2)$ must be equal to 0. Let's substitute $x = 2$ into the polynomial:

• $p(2) = (2)^3 - a(2)^2 + b(2) + 6$
• $0 = 8 - a(4) + 2b + 6$
• $0 = 8 - 4a + 2b + 6$
• $0 = 14 - 4a + 2b$
• We can simplify this equation by dividing by 2: $0 = 7 - 2a + b$
• This gives us our second equation: $2a - b = 7$ (Equation 2)

Solving the System of Equations

Now we have a system of two linear equations with two variables $a$ and $b$:

Equation 1: $a + b = 5$

Equation 2: $2a - b = 7$

We can solve this system using the elimination method by adding the two equations:

$(a + b) + (2a - b) = 5 + 7$

$a + b + 2a - b = 12$

$3a = 12$

Now, solve for $a$:

$a = \frac{12}{3}$

$a = 4$

Now that we have the value of $a$, we can substitute it back into either Equation 1 or Equation 2 to find the value of $b$. Using Equation 1:

$a + b = 5$

$4 + b = 5$

$b = 5 - 4$

$b = 1$

Thus, the respective values of $a$ and $b$ are 4 and 1.

Let's verify this using Equation 2: $2a - b = 2(4) - 1 = 8 - 1 = 7$. This confirms our values are correct.

The values are $a=4$ and $b=1$. These are the respective values as requested.