Simplification!

Hello, Pandp77!

Quote:

$p(x) \:= \:ax^3 + bx^2 - 5x + c$ is a polynomial exactly divisible by $(x+2).$

Also: . $p(0) \:= \:p(1) \:= \:-6$

Find $a, b\text{ and }c$ , and completely factorize $p(x)$

Since $p(x)$ is divisible by $(x+2)$ , then: . $f(\text{-}2) \:=\:0$
. . $a(\text{-}2)^3 + b(\text{-}2)^2 - 5(\text{-}2) + c \:=\:0 \quad\Rightarrow\quad \text{-}8a + 4b + c \:=\:\text{-}10\;\;{\color{blue}[1]}$

Since $p(0) = \text{-}6\!:\;\;a(0^3) + b(0^2) - 5(0) + c \:=\:\text{-}6\quad\Rightarrow\quad\boxed{ c \:=\:-6}\;\;{\color{blue}[2]}$

Since $p(1) = \text{-}6\!:\;\;a(1^3) + b(1^2) - 5(1) + c \:=\:\text{-}6\quad\Rightarrow\quad a + b + c \:=\:\text{-}1\;\;{\color{blue}[3]}$

Substitute [2] into [1]: . $\text{-}8a + 4b -6 \:=\:\text{-}10 \quad\Rightarrow\quad 2a - b \:=\:1\;\;{\color{blue}[4]}$

Substitute [2] into [3]: . . $a + b -6 \:=\:-1\quad\Rightarrow\quad\;\; a + b \:=\:5\;\;{\color{blue}[5]}$

Add [4] and [5]: . $3a \:=\:6 \quad\Rightarrow\quad \boxed{a \:=\:2}$

Substitute into [5]: . $2 + b \:=\:5\quad\Rightarrow\quad\boxed{ b \:=\:3}$

Hence: . $p(x) \;=\;2x^3 + 3x^2 - 5x - 6$

We know that $p(x)$ is divisible by $(x+2)\!: \;\;p(x) \;=\;(x+2)(2x^2 - x - 3)$

Therefore: . $p(x) \;=\;(x+2)(x+1)(2x-3)$