Sure! Let's solve the problem step by step.
Given the polynomial function [tex]\( f(x) = 2x^3 + ax^2 + bx - 6 \)[/tex] and the conditions that the remainder is 12 when [tex]\( f(x) \)[/tex] is divided by [tex]\((x-1)\)[/tex] and -18 when it is divided by [tex]\((x+4)\)[/tex].
### Step 1: Using the Remainder Theorem
The Remainder Theorem states that if a polynomial [tex]\( f(x) \)[/tex] is divided by [tex]\( x - c \)[/tex], the remainder of that division is [tex]\( f(c) \)[/tex].
#### Condition 1: [tex]\( f(1) = 12 \)[/tex]
Substitute [tex]\( x = 1 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(1) = 2(1)^3 + a(1)^2 + b(1) - 6 \][/tex]
[tex]\[ 12 = 2 + a + b - 6 \][/tex]
[tex]\[ 12 = a + b - 4 \][/tex]
[tex]\[ a + b = 16 \][/tex]
We have our first equation:
[tex]\[ a + b = 16 \quad \text{(Equation 1)} \][/tex]
#### Condition 2: [tex]\( f(-4) = -18 \)[/tex]
Substitute [tex]\( x = -4 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-4) = 2(-4)^3 + a(-4)^2 + b(-4) - 6 \][/tex]
[tex]\[ -18 = 2(-64) + a(16) + b(-4) - 6 \][/tex]
[tex]\[ -18 = -128 + 16a - 4b - 6 \][/tex]
[tex]\[ -18 = 16a - 4b - 134 \][/tex]
[tex]\[ 16a - 4b = 116 \][/tex]
Simplify this equation:
[tex]\[ 4a - b = 29 \][/tex]
We have our second equation:
[tex]\[ 4a - b = 29 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solving the System of Equations
Now we solve the system of linear equations:
[tex]\[ \begin{cases}
a + b = 16 \quad \text{(Equation 1)} \\
4a - b = 29 \quad \text{(Equation 2)}
\end{cases} \][/tex]
Add the two equations to eliminate [tex]\( b \)[/tex]:
[tex]\[ (a + b) + (4a - b) = 16 + 29 \][/tex]
[tex]\[ a + 4a + b - b = 45 \][/tex]
[tex]\[ 5a = 45 \][/tex]
[tex]\[ a = 9 \][/tex]
Now substitute [tex]\( a = 9 \)[/tex] back into Equation 1:
[tex]\[ 9 + b = 16 \][/tex]
[tex]\[ b = 16 - 9 \][/tex]
[tex]\[ b = 7 \][/tex]
### Conclusion
Thus, the values of the constants are:
[tex]\[ a = 9 \][/tex]
[tex]\[ b = 7 \][/tex]
So, the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are [tex]\( 9 \)[/tex] and [tex]\( 7 \)[/tex], respectively.
