Let's tackle the problem step-by-step to find the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex] and then solve for the inequality [tex]\( p(x) < 0 \)[/tex].
### Part (a): Determine [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
Given:
1. The polynomial [tex]\( p(x) = a x^4 + 3 x^3 + 2 x^2 + 6 x + b \)[/tex]
2. It is divisible by [tex]\( 2x - 1 \)[/tex]. Since it's divisible by [tex]\( 2x - 1 \)[/tex], when [tex]\( x = \frac{1}{2} \)[/tex], [tex]\( p \left( \frac{1}{2} \right) = 0 \)[/tex].
3. The remainder when [tex]\( p(x) \)[/tex] is divided by [tex]\( x - 1 \)[/tex] is 9. So, [tex]\( p(1) = 9 \)[/tex].
Finding [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
1. Using [tex]\( p \left( \frac{1}{2} \right) = 0 \)[/tex]:
Substitute [tex]\( x = \frac{1}{2} \)[/tex] into the polynomial:
[tex]\[
p \left( \frac{1}{2} \right) = a \left( \frac{1}{2} \right)^4 + 3 \left( \frac{1}{2} \right)^3 + 2 \left( \frac{1}{2} \right)^2 + 6 \left( \frac{1}{2} \right) + b = 0
\][/tex]
Simplify the equation:
[tex]\[
a \left( \frac{1}{16} \right) + 3 \left( \frac{1}{8} \right) + 2 \left( \frac{1}{4} \right) + 6 \left( \frac{1}{2} \right) + b = 0
\][/tex]
[tex]\[
\frac{a}{16} + \frac{3}{8} + \frac{2}{4} + 3 + b = 0
\][/tex]
Convert all terms to have a common denominator:
[tex]\[
\frac{a}{16} + \frac{6}{16} + \frac{8}{16} + \frac{48}{16} + b = 0
\][/tex]
Combine like terms:
[tex]\[
\frac{a + 6 + 8 + 48}{16} + b = 0
\][/tex]
[tex]\[
\frac{a + 62}{16} + b = 0
\][/tex]
Multiply through by 16:
[tex]\[
a + 62 + 16b = 0
\][/tex]
[tex]\[
a + 16b = -62 \quad \text{ (Equation 1)}
\][/tex]
2. Using [tex]\( p(1) = 9 \)[/tex]:
Substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[
p(1) = a(1)^4 + 3(1)^3 + 2(1)^2 + 6(1) + b = 9
\][/tex]
Simplify the equation:
[tex]\[
a + 3 + 2 + 6 + b = 9
\][/tex]
[tex]\[
a + b + 11 = 9
\][/tex]
[tex]\[
a + b = -2 \quad \text{ (Equation 2)}
\][/tex]
Solving the system of equations:
[tex]\[
\begin{cases}
a + 16b = -62 \quad \text{ (Equation 1)} \\
a + b = -2 \quad \text{ (Equation 2)}
\end{cases}
\][/tex]
Subtract Equation 2 from Equation 1:
[tex]\[
(a + 16b) - (a + b) = -62 - (-2)
\][/tex]
[tex]\[
15b = -60
\][/tex]
[tex]\[
b = -4
\][/tex]
Substitute [tex]\( b = -4 \)[/tex] into Equation 2:
[tex]\[
a + (-4) = -2
\][/tex]
[tex]\[
a - 4 = -2
\][/tex]
[tex]\[
a = 2
\][/tex]
Thus, the constants are:
[tex]\[
a = 2, \ b = -4
\][/tex]
### Part (b): Find the solution set for the inequality [tex]\( p(x) < 0 \)[/tex]
Now substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] back into the polynomial:
[tex]\[
p(x) = 2x^4 + 3x^3 + 2x^2 + 6x - 4
\][/tex]
To solve the inequality [tex]\( p(x) < 0 \)[/tex], we need to find the roots of the polynomial [tex]\( p(x) \)[/tex] and analyze the intervals.
The roots of the polynomial (identifying the roots exactly is complex, so using numerical solutions gives us):
[tex]\[
(x + 2)(other \ factors)
\][/tex]
From solving the inequality [tex]\( p(x) < 0 \)[/tex]:
The solution is found:
[tex]\[
\boxed{(-2 < x < 0.5)}
\][/tex]
Therefore, the solution set for the inequality [tex]\( p(x) < 0 \)[/tex] is [tex]\( x \)[/tex] in the interval [tex]\( -2 < x < 0.5 \)[/tex].
