use the remainder theorem to find $p(3)$ for $p(x)=-2x^{3}+4x^{2}-x-5$.\nspecifically, give the quotient and…

use the remainder theorem to find $p(3)$ for $p(x)=-2x^{3}+4x^{2}-x-5$.\nspecifically, give the quotient and the remainder for the associated division and the value of $p(3)$.\nquotient = $\\square$\nremainder = $\\square$\n$p(3) = \\square$
Answer
Explanation:
Step1: Set up polynomial division
Divide $P(x)=-2x^3 + 4x^2 - x - 5$ by $x-3$ (since we find $P(3)$).
Step2: Divide leading terms
$\frac{-2x^3}{x}=-2x^2$. Multiply $x-3$ by $-2x^2$: $-2x^3+6x^2$. Subtract from $P(x)$: $$(-2x^3 + 4x^2 - x - 5)-(-2x^3+6x^2)=-2x^2 - x - 5$$
Step3: Divide new leading terms
$\frac{-2x^2}{x}=-2x$. Multiply $x-3$ by $-2x$: $-2x^2+6x$. Subtract: $$(-2x^2 - x - 5)-(-2x^2+6x)=-7x - 5$$
Step4: Divide next leading terms
$\frac{-7x}{x}=-7$. Multiply $x-3$ by $-7$: $-7x+21$. Subtract: $$(-7x - 5)-(-7x+21)=-26$$
Step5: Apply Remainder Theorem
By Remainder Theorem, $P(3)$ equals the remainder of the division.
Answer:
Quotient = $-2x^2 - 2x - 7$ Remainder = $-26$ $P(3) = -26$