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.

$\displaystyle (x-1)(x+3)(-x+p)$
=($\displaystyle x^2$+2x-3)(-x+p)
=$\displaystyle -x^3$$\displaystyle +x^2$$\displaystyle p$$\displaystyle -2x^2$$\displaystyle +2xp+3x-3p$

let x be -4,
$\displaystyle -(-4)^3$ $\displaystyle + (-4)^2$$\displaystyle p$$\displaystyle -2(-4)^2$$\displaystyle + 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, $\displaystyle f(x) = 0$, are $\displaystyle 1,-3\text{ and }p.$
Given that $\displaystyle f(x)$ is such that the coefficient of $\displaystyle x^3$ is -1,
and that it has a remainder of 10 when divided by $\displaystyle (x+4)$, find:

(a) the value of $\displaystyle p.$

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

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

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

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

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

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

Answer: .$\displaystyle (ii)$

3. Originally Posted by Soroban
Hello, wintersoltice!

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

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

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

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

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

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

Answer: .$\displaystyle (ii)$

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.