Answer :
Answer:
No
Step-by-step explanation:
We can determine whether (x + 7) is a factor of the cubic function:
[tex]f(x)=x^3-3x^2+2x-8[/tex]
using polynomial long division:
[tex]\text{ }\ \ \ \ \ \ \ \ \ x^2-10x+72\\ x+7 \ )\!\overline{\ x^3-3x^2+2x-8} \\ \text{ }\ \ \ \ \ \underline{-(x^3 + 7x^2)} \\ \text{ }\ \ \ \ \ \ \ \ \: 0-10x^2 + 2x \\ \text{ }\ \ \ \ \ \ \ \ \ \ \underline{-(10x^2 - 70x)} \\ \text{ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 + 72x - 8 \\ \text{ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-(72x+504)} \\ \text{ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R \ \text{-}\, 512[/tex]
Since there is a remainder of -512, (x + 7) is NOT a factor of the cubic function.
Further Note
We can technically write (x + 7) as a factor of the function, but it requires (x + 7) also being in the denominator of a fraction, thus canceling itself when multiplied out:
[tex]f(x)=\left(x+7\right)\!\left(x^2-10x+72-\dfrac{512}{x+7}\right)[/tex]
