First, we use the information given about the remainders when [tex]\(P(z)\)[/tex] is divided by [tex]\(z+1\)[/tex] and [tex]\(z-3\)[/tex]. The Remainder Theorem tells us that for a polynomial [tex]\(P(z)\)[/tex]:
1. When [tex]\(P(z)\)[/tex] is divided by [tex]\(z+1\)[/tex], the remainder is [tex]\(P(-1) = -8\)[/tex].
2. When [tex]\(P(z)\)[/tex] is divided by [tex]\(z-3\)[/tex], the remainder is [tex]\(P(3) = 4\)[/tex].
When [tex]\(P(z)\)[/tex] is divided by [tex]\((z-3)(z+1)\)[/tex], the remainder will be a linear polynomial of the form [tex]\(R(z) = az + b\)[/tex], because the degree of the remainder must be less than the degree of the divisor [tex]\((z-3)(z+1)\)[/tex], which is quadratic.
Therefore, we can write:
[tex]\[ P(z) = Q(z) \cdot (z-3)(z+1) + az + b \][/tex]
for some polynomial [tex]\(Q(z)\)[/tex].
To find the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we substitute [tex]\(z = -1\)[/tex] and [tex]\(z = 3\)[/tex] into the polynomial.
1. Substitute [tex]\(z = -1\)[/tex]:
[tex]\[ P(-1) = a(-1) + b = -a + b \][/tex]
Given that [tex]\(P(-1) = -8\)[/tex], we have the equation:
[tex]\[ -a + b = -8 \quad \text{(Equation 1)} \][/tex]
2. Substitute [tex]\(z = 3\)[/tex]:
[tex]\[ P(3) = 3a + b \][/tex]
Given that [tex]\(P(3) = 4\)[/tex], we have the equation:
[tex]\[ 3a + b = 4 \quad \text{(Equation 2)} \][/tex]
Now we solve this system of linear equations:
From Equation 1:
[tex]\[ -a + b = -8 \][/tex]
From Equation 2:
[tex]\[ 3a + b = 4 \][/tex]
Subtract Equation 1 from Equation 2 to eliminate [tex]\(b\)[/tex]:
[tex]\[ (3a + b) - (-a + b) = 4 - (-8) \][/tex]
[tex]\[ 3a + b + a - b = 4 + 8 \][/tex]
[tex]\[ 4a = 12 \][/tex]
[tex]\[ a = 3 \][/tex]
Substitute [tex]\(a = 3\)[/tex] back into Equation 1 to find [tex]\(b\)[/tex]:
[tex]\[ -3 + b = -8 \][/tex]
[tex]\[ b = -8 + 3 \][/tex]
[tex]\[ b = -5 \][/tex]
Therefore, the remainder [tex]\(R(z)\)[/tex] when [tex]\(P(z)\)[/tex] is divided by [tex]\((z-3)(z+1)\)[/tex] is:
[tex]\[ R(z) = 3z - 5 \][/tex]
