Certainly! To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] so that [tex]\(x+1\)[/tex] and [tex]\(x-1\)[/tex] are factors of the polynomial [tex]\(x^4 + a x^3 + 2 x^2 - 3 x + b\)[/tex], we need to follow these steps:
1. Apply the Factor Theorem: The Factor Theorem states that [tex]\(x - c\)[/tex] is a factor of a polynomial if and only if the polynomial evaluated at [tex]\(c\)[/tex] is zero. Therefore, [tex]\(x+1\)[/tex] and [tex]\(x-1\)[/tex] being factors implies:
- [tex]\(P(-1) = 0\)[/tex]
- [tex]\(P(1) = 0\)[/tex]
2. Set up the polynomial [tex]\(P(x)\)[/tex]:
[tex]\[
P(x) = x^4 + a x^3 + 2 x^2 - 3 x + b
\][/tex]
3. Evaluate the polynomial at [tex]\(x = -1\)[/tex] and [tex]\(x = 1\)[/tex]:
- For [tex]\(x = -1\)[/tex]:
[tex]\[
P(-1) = (-1)^4 + a(-1)^3 + 2(-1)^2 - 3(-1) + b = 0
\][/tex]
Simplifying the above:
[tex]\[
1 - a + 2 + 3 + b = 0 \implies 6 - a + b = 0 \implies a - b = 6 \quad \text{(Equation 1)}
\][/tex]
- For [tex]\(x = 1\)[/tex]:
[tex]\[
P(1) = (1)^4 + a(1)^3 + 2(1)^2 - 3(1) + b = 0
\][/tex]
Simplifying the above:
[tex]\[
1 + a + 2 - 3 + b = 0 \implies a + b = 0 \quad \text{(Equation 2)}
\][/tex]
4. Solve the system of linear equations obtained:
We have:
[tex]\[
\begin{cases}
a - b = 6 & \quad \text{(Equation 1)} \\
a + b = 0 & \quad \text{(Equation 2)}
\end{cases}
\][/tex]
To solve these, we can add both equations:
[tex]\[
(a - b) + (a + b) = 6 + 0 \implies 2a = 6 \implies a = 3
\][/tex]
Substitute [tex]\(a = 3\)[/tex] into Equation 2:
[tex]\[
3 + b = 0 \implies b = -3
\][/tex]
Therefore, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that make [tex]\(x+1\)[/tex] and [tex]\(x-1\)[/tex] factors of the polynomial [tex]\(x^4 + a x^3 + 2 x^2 - 3 x + b\)[/tex] are:
[tex]\[
a = 3 \quad \text{and} \quad b = -3
\][/tex]
