1. ## another question about factor theorem/polynomials

Question: please refer to the picture above.
I think there is something wrong about my working as my remainder's answer is different from the choices, but i don't see anywhere went wrong. ="=
Mind helping me to check whether is there anything wrong in my working?

NOTE: I don't really know how to use [ math ] [ /math]. So, if you see some small p or big p, they're the same.

$
(x-1)(x+3)(-x+p)$

=( $x^2$+2x-3)(-x+p)
= $-x^3$ $+x^2$ $p$ $-2x^2$ $+2xp+3x-3p$

let x be -4,
$-(-4)^3$ $+ (-4)^2$ $p$ $-2(-4)^2$ $+ 2(-4)p +3(-4) -3p=10$
64 + 16p -32 -8p-12-3p=10
5p=-10
p=-2

let x be 4 and substitute p=-2,
-64 -32 -16 -16 +12 +6 = -110
my remainder is -110, but according to the choices, the remainder should be -126.

2. Hello, wintersoltice!

The roots of a cubic equationj, $f(x) = 0$, are $1,-3\text{ and }p.$
Given that $f(x)$ is such that the coefficient of $x^3$ is -1,
and that it has a remainder of 10 when divided by $(x+4)$, find:

(a) the value of $p.$

(b) the remainder when divided by $(x-4)$
The function is: . $f(x) \:=\:-(x-1)(x+3)(x-p)$

(a) "The remainder is 10 when divided by $(x+4)$": . $f(\text{-}4) = 10$

. . $f(\text{-}4) \:=\:-(\text{-}4-1)(\text{-}4+3)(\text{-}4-p) \:=\:10 \quad\Rightarrow\quad 5(p+4) \:=\:10 \quad\Rightarrow\quad \boxed{p \:=\:-2}$

(b) The function is: . $f(x) \:=\:-(x-1)(x+3)(x+2)$

"The remainder when $f(x)$ is divided by $(x-4)$": .we want $f(4)$

. . $f(4) \:=\:-(4-1)(4+3)(4+2) \:=\:\boxed{-126}$

Answer: . $(ii)$

3. Originally Posted by Soroban
Hello, wintersoltice!

The function is: . $f(x) \:=\:-(x-1)(x+3)(x-p)$

(a) "The remainder is 10 when divided by $(x+4)$": . $f(\text{-}4) = 10$

. . $f(\text{-}4) \:=\:-(\text{-}4-1)(\text{-}4+3)(\text{-}4-p) \:=\:10 \quad\Rightarrow\quad 5(p+4) \:=\:10 \quad\Rightarrow\quad \boxed{p \:=\:-2}$

(b) The function is: . $f(x) \:=\:-(x-1)(x+3)(x+2)$

"The remainder when $f(x)$ is divided by $(x-4)$": .we want $f(4)$

. . $f(4) \:=\:-(4-1)(4+3)(4+2) \:=\:\boxed{-126}$

Answer: . $(ii)$

o i see.
so, my p is correct.
but i still don't know where i went wrong in my working, making my remainder incorrect. =_=
oh well, anyway, thanks. =)

4. Originally Posted by wintersoltice
[snip]
let x be 4 and substitute p=-2,
-64 -32 -16 -16 +12 +6 = -110
my remainder is -110, but according to the choices, the remainder should be -126.
Careless arithmetic. It should be -64 - 32 - 32 - 16 + 12 + 6.