Show that $(x + 4), (x - 3)$ and $(x - 7)$ are factors of $x^3 - 6x^2 - 19x + 84$.

Given:

Given polynomial is $x^3 - 6x^2 - 19x + 84$.

To do:

We have to show that $(x + 4), (x - 3)$ and $(x - 7)$ are factors of $x^3 - 6x^2 - 19x + 84$.

Solution:

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

To check whether $(x + 4), (x - 3)$ and $(x - 7)$ are factors of $x^3 - 6x^2 - 19x + 84$, we have to substitue $x=-4, x=3$ and $x=7$ respectively in $x^3 - 6x^2 - 19x + 84$.

Let $f(x) = x^3 - 6x^2 - 19x + 84$

$f(-4) = (-4)^3-6(-4)^2 -19(-4)+84$

$= -64-96+76+84$

$=160-160$

$=0$

$f(3)=(3)^3-6(3)^2 -19(3)+84$

$= 27-6(9)-57+84$

$=111-111$

$=0$

$f(7) = (7)^3-6(7)^2 -19(7)+84$

$= 343-294-133+84$

$=427-427$

$=0$

This implies, $(x+4), (x-3)$ and $(x-7)$ are factors of $x^3 - 6x^2 - 19x + 84$.

Hence proved. 

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Updated on: 10-Oct-2022

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