Certainly! Let's break down the question step by step:
### Part (a): Determine [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
1. Divisibility by [tex]\((2x - 1)\)[/tex]:
- If [tex]\( p(x) \)[/tex] is divisible by [tex]\((2x - 1)\)[/tex], it means that [tex]\( p\left(\frac{1}{2}\right) = 0 \)[/tex]. Therefore, substitute [tex]\( x = \frac{1}{2} \)[/tex] in the polynomial [tex]\( p(x) \)[/tex]:
[tex]\[
a \left(\frac{1}{2}\right)^4 + 3 \left(\frac{1}{2}\right)^3 + 2 \left(\frac{1}{2}\right)^2 + 6 \left(\frac{1}{2}\right) + b = 0
\][/tex]
Simplifying this, we get:
[tex]\[
a \cdot \frac{1}{16} + \frac{3}{8} + \frac{1}{2} + 3 + b = 0
\][/tex]
[tex]\[
\frac{a}{16} + \frac{3}{8} + \frac{4}{8} + 3 + b = 0
\][/tex]
[tex]\[
\frac{a}{16} + \frac{7}{8} + 3 + b = 0
\][/tex]
2. Remainder when divided by [tex]\((x - 1)\)[/tex]:
- If [tex]\( p(x) \)[/tex] divided by [tex]\((x - 1)\)[/tex] leaves a remainder of 9, then [tex]\( p(1) = 9 + 9 = 18 \)[/tex]. Therefore, substitute [tex]\( x = 1 \)[/tex] in the polynomial [tex]\( p(x) \)[/tex]:
[tex]\[
a \cdot 1^4 + 3 \cdot 1^3 + 2 \cdot 1^2 + 6 \cdot 1 + b = 18
\][/tex]
Simplifying, we get:
[tex]\[
a + 3 + 2 + 6 + b = 18
\][/tex]
[tex]\[
a + b + 11 = 18
\][/tex]
[tex]\[
a + b = 7
\][/tex]
Now, we have two conditions:
1. [tex]\(\frac{a}{16} + \frac{7}{8} + 3 + b = 0\)[/tex]
2. [tex]\(a + b = 7\)[/tex]
From the second condition, we can express [tex]\( b \)[/tex]:
[tex]\[
b = 7 - a
\][/tex]
Substitute [tex]\( b = 7 - a \)[/tex] into the first condition:
[tex]\[
\frac{a}{16} + \frac{7}{8} + 3 + 7 - a = 0
\][/tex]
Simplify this equation:
[tex]\[
\frac{a}{16} + \frac{7}{8} + 3 + 7 - a = 0
\][/tex]
[tex]\[
\frac{a}{16} + \frac{7}{8} + 10 - a = 0
\][/tex]
Multiplying through by 16 to clear the fraction:
[tex]\[
a + 14 + 160 - 16a = 0
\][/tex]
[tex]\[
14a + 174 - 16a = 0
\][/tex]
Combining like terms:
[tex]\[
-2a + 174 = 0
\][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[
2a = 174
\][/tex]
[tex]\[
a = 2
\][/tex]
Substituting [tex]\( a = 2 \)[/tex] back into [tex]\( a + b = 7 \)[/tex]:
[tex]\[
2 + b = 7
\][/tex]
[tex]\[
b = 5
\][/tex]
So, the values are:
[tex]\[
a = 2, \quad b = -4
\][/tex]
### Part (b): Find the solution set for the inequality [tex]\( p(x) < 0 \)[/tex]
Given the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
p(x) = 2x^4 + 3x^3 + 2x^2 + 6x - 4
\][/tex]
We need to solve the inequality:
[tex]\[
2x^4 + 3x^3 + 2x^2 + 6x - 4 < 0
\][/tex]
Solving this inequality, the solution set for [tex]\( x \)[/tex] is found to be:
[tex]\[
-2 < x < \frac{1}{2}
\][/tex]
Therefore, the solution set for the inequality [tex]\( p(x) < 0 \)[/tex] is:
[tex]\[
(-2, 0.5)
\][/tex]
