To find the constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in the polynomial [tex]\(P(x) = x^3 + 2x^2 + ax + b\)[/tex], given the remainders when divided by [tex]\(x-1\)[/tex] and [tex]\(x+2\)[/tex], follow these steps:
1. Remainder when divided by [tex]\(x-1\)[/tex]:
When [tex]\(P(x)\)[/tex] is divided by [tex]\(x-1\)[/tex], the remainder is given as 4. We can use the Remainder Theorem to find:
[tex]\[
P(1) = 1^3 + 2(1)^2 + a(1) + b = 4
\][/tex]
Simplifying, we get:
[tex]\[
1 + 2 + a + b = 4 \implies 3 + a + b = 4 \implies a + b = 1 \quad \text{(Equation 1)}
\][/tex]
2. Remainder when divided by [tex]\(x+2\)[/tex]:
When [tex]\(P(x)\)[/tex] is divided by [tex]\(x+2\)[/tex], the remainder is given as 16. Again, using the Remainder Theorem, we find:
[tex]\[
P(-2) = (-2)^3 + 2(-2)^2 + a(-2) + b = 16
\][/tex]
Simplifying, we get:
[tex]\[
-8 + 8 - 2a + b = 16 \implies 0 - 2a + b = 16 \implies -2a + b = 16 \quad \text{(Equation 2)}
\][/tex]
3. Solving the system of equations:
Now we solve the system of linear equations:
[tex]\[
\begin{cases}
a + b = 1 \quad &\text{(Equation 1)} \\
-2a + b = 16 \quad &\text{(Equation 2)}
\end{cases}
\][/tex]
Subtract Equation 1 from Equation 2 to eliminate [tex]\(b\)[/tex]:
[tex]\[
(-2a + b) - (a + b) = 16 - 1
\][/tex]
Simplifying, we get:
[tex]\[
-2a + b - a - b = 15 \implies -3a = 15 \implies a = -5
\][/tex]
Substitute [tex]\(a = -5\)[/tex] back into Equation 1:
[tex]\[
-5 + b = 1 \implies b = 6
\][/tex]
Hence, the constants are:
[tex]\[
a = -5, \quad b = 6
\][/tex]
