To determine the values of [tex]\( a \)[/tex] for which [tex]\( (x-a) \)[/tex] is a factor of the polynomial [tex]\( a x^3 - 3 x^2 - 5 a x - 9 \)[/tex], we must ensure that substituting [tex]\( x = a \)[/tex] into the polynomial yields zero. This implies that [tex]\( a \)[/tex] is a root of the polynomial when substituting [tex]\( x = a \)[/tex].
1. Substitute [tex]\( x = a \)[/tex] into the polynomial:
[tex]\[
a \cdot a^3 - 3 \cdot a^2 - 5a \cdot a - 9 = 0
\][/tex]
2. Simplify the equation:
[tex]\[
a^4 - 3a^2 - 5a^2 - 9 = 0
\][/tex]
[tex]\[
a^4 - 8a^2 - 9 = 0
\][/tex]
3. Solve for [tex]\( a \)[/tex]:
This is a quadratic equation in terms of [tex]\( a^2 \)[/tex]. Let [tex]\( z = a^2 \)[/tex]. Then:
[tex]\[
z^2 - 8z - 9 = 0
\][/tex]
Using the quadratic formula, [tex]\( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = -9 \)[/tex]:
[tex]\[
z = \frac{8 \pm \sqrt{64 + 36}}{2}
\][/tex]
[tex]\[
z = \frac{8 \pm 10}{2}
\][/tex]
Thus, we have two solutions for [tex]\( z \)[/tex]:
[tex]\[
z = 9 \quad \text{or} \quad z = -1
\][/tex]
Since [tex]\( z = a^2 \)[/tex], we find the values of [tex]\( a \)[/tex]:
[tex]\[
a^2 = 9 \implies a = 3 \quad \text{or} \quad a = -3
\][/tex]
[tex]\[
a^2 = -1 \implies a = i \quad \text{or} \quad a = -i
\][/tex]
Thus, the possible values for [tex]\( a \)[/tex] are [tex]\( 3 \)[/tex], [tex]\( -3 \)[/tex], [tex]\( i \)[/tex], and [tex]\( -i \)[/tex].
4. Factorize the expression for each value of [tex]\( a \)[/tex]:
- For [tex]\( a = 3 \)[/tex]:
[tex]\[
3 x^3 - 3 x^2 - 15 x - 9
\][/tex]
This polynomial can be factored as:
[tex]\[
3(x - 1)^2(x + 3)
\][/tex]
- For [tex]\( a = -3 \)[/tex]:
[tex]\[
-3 x^3 - 3 x^2 + 15 x - 9
\][/tex]
This polynomial can be factored as:
[tex]\[
-3(x - 1)^2(x + 3)
\][/tex]
- For [tex]\( a = i \)[/tex]:
[tex]\[
i x^3 - 3 x^2 - 5 i x - 9
\][/tex]
This polynomial can be factored as:
[tex]\[
i(x - i)(x^2 + 4ix + 9)
\][/tex]
- For [tex]\( a = -i \)[/tex]:
[tex]\[
-i x^3 - 3 x^2 + 5 i x - 9
\][/tex]
This polynomial can be factored as:
[tex]\[
-i(x + i)(x^2 - 4ix + 9)
\][/tex]
To summarize, the possible values of [tex]\( a \)[/tex] are [tex]\( 3 \)[/tex], [tex]\( -3 \)[/tex], [tex]\( i \)[/tex], and [tex]\( -i \)[/tex]. The factorizations are:
- For [tex]\( a = 3 \)[/tex]: [tex]\( 3(x - 1)^2(x + 3) \)[/tex]
- For [tex]\( a = -3 \)[/tex]: [tex]\( -3(x - 1)^2(x + 3) \)[/tex]
- For [tex]\( a = i \)[/tex]: [tex]\( i(x - i)(x^2 + 4ix + 9) \)[/tex]
- For [tex]\( a = -i \)[/tex]: [tex]\( -i(x + i)(x^2 - 4ix + 9) \)[/tex]
