To find the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex] for which the expression [tex]\( x = 4x^4 - 6x^2 + px + q \)[/tex] is divisible by [tex]\( x^2 - 1 \)[/tex], we need to ensure that the expression does not leave any remainder when divided by [tex]\( x^2 - 1 \)[/tex].
The polynomial [tex]\( x^2 - 1 \)[/tex] can be factored as [tex]\( (x - 1)(x + 1) \)[/tex]. Therefore, for [tex]\( 4x^4 - 6x^2 + px + q \)[/tex] to be divisible by [tex]\( x^2 - 1 \)[/tex], it must yield zero when [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex]. This gives us the following steps:
1. Substitute [tex]\( x = 1 \)[/tex] into the expression:
[tex]\[
4(1)^4 - 6(1)^2 + p(1) + q = 0
\][/tex]
Simplifying this, we get:
[tex]\[
4 - 6 + p + q = 0
\][/tex]
[tex]\[
p + q - 2 = 0
\][/tex]
This simplifies to our first equation:
[tex]\[
p + q = 2 \quad \text{(Equation 1)}
\][/tex]
2. Substitute [tex]\( x = -1 \)[/tex] into the expression:
[tex]\[
4(-1)^4 - 6(-1)^2 + p(-1) + q = 0
\][/tex]
Simplifying this, we get:
[tex]\[
4 - 6 - p + q = 0
\][/tex]
[tex]\[
-p + q - 2 = 0
\][/tex]
This simplifies to our second equation:
[tex]\[
-p + q = 2 \quad \text{(Equation 2)}
\][/tex]
3. Solve the system of simultaneous linear equations:
[tex]\[
\begin{cases}
p + q = 2 \\
-p + q = 2
\end{cases}
\][/tex]
- Add the two equations:
[tex]\[
(p + q) + (-p + q) = 2 + 2
\][/tex]
[tex]\[
q + q = 4
\][/tex]
[tex]\[
2q = 4
\][/tex]
[tex]\[
q = 2
\][/tex]
- Substitute [tex]\( q = 2 \)[/tex] back into Equation 1:
[tex]\[
p + 2 = 2
\][/tex]
[tex]\[
p = 0
\][/tex]
Therefore, the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex] which make [tex]\( 4x^4 - 6x^2 + px + q \)[/tex] divisible by [tex]\( x^2 - 1 \)[/tex] without leaving a remainder are:
[tex]\[
p = 0 \quad \text{and} \quad q = 2
\][/tex]
