Show that $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24$.

Given:

Given polynomial is $x^3 - 3x^2 - 10x + 24$.

To do:

We have to show that $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24$.

Solution:

We know that, if $g(x)$ is a factor of $f(x)$, then the remainder will be zero.

To check whether $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24$, we have to substitue $x=2, x=-3$ and $x=4$ respectively in $x^3 - 3x^2 - 10x + 24$.

Let $f(x) = x^3 - 3x^2 - 10x + 24$

$f(2) = (2)^3-3(2)^2 -10(2)+24$

$= 8-12-20+24$

$=32-32$

$=0$

$f(-3)=(-3)^3-3(-3)^2 -10(-3)+24$

$= -27-3(9)+30+24$

$=54-54$

$=0$

$f(4) = (4)^3-3(4)^2 -10(4)+24$

$= 64-48-40+24$

$=88-88$

$=0$

This implies, $(x-2), (x+3)$ and $(x-4)$ are factors of $x^3 - 3x^2 - 10x + 24$.

Hence proved.

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Updated on: 10-Oct-2022

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