Answer: [tex]\begin{gathered} Quotient\text{ = 2x}^2 \\ Remainder\text{ = -8} \\ P(3)\text{ = -8} \end{gathered}[/tex]
Explanation:[tex]\begin{gathered} Given\text{ polynomial:} \\ P(x)\text{ = 2x}^3\text{ - 6x}^2\text{ - 8} \\ We\text{ need to apply the remainder theorem to get the quotient and remainder} \end{gathered}[/tex]
P(3): To apply the remainder theorem, we will substitute the value of x with 3 in the function
[tex]\begin{gathered} P(3)\text{ = 2\lparen3\rparen}^3\text{ - 6\lparen3\rparen}^2\text{ - 8} \\ P(3)\text{ = 54 - 54 - 8} \\ P(3)\text{ = -8} \end{gathered}[/tex][tex]\begin{gathered} The\text{ remainder is value left after substituting the value for x} \\ The\text{ remainder = -8} \end{gathered}[/tex]
To get the quotient, we will apply long division:
[tex]\begin{gathered} From\text{ }P(3),\text{ x = 3} \\ \text{ gives a factor of \lparen x - 3\rparen} \\ \frac{2x^3-6x^2-8}{(x-3)} \end{gathered}[/tex][tex]\begin{gathered} Quotient\text{ = 2x}^2 \\ Remainder\text{ = -8} \\ P(3)\text{ = -8} \end{gathered}[/tex]
