# Find the factors of an equation

zak100

## Homework Statement

I have following eq:

(x)^3 -12(x) -16=0

How to find x-4 a factor of this eq. This means we have:
(x-4) (x^2 + 4x +4) =0

I don’t know how to get the above.

## Homework Equations

(x)^3 -12(x) -16=0

## The Attempt at a Solution

I cant go beyond that:
(x)^3 -12(x) -16=0
X(x^2 -12-16/x) = 0

Last edited by a moderator:

## Answers and Replies

Mentor
How to find x-4 a factor of this eq.
What's an x-4 factor? Do you mean that you want to factor the original equation to solve it for values of x that satisfy the equation?

zak100
Hi,
If we put x=4 then

(x)^3 -12(x) -16=0 becomes
(4)^3 -12(4) -16 =0
64 -48 -16 =0
So its zero.

So x =4 &
x-4=0
but how to get:

(x^2 + 4x +4) =0

Zulfi.

Staff Emeritus
Homework Helper
Gold Member
Hi,
If we put x=4 then

(x)^3 -12(x) -16=0 becomes
(4)^3 -12(4) -16 =0
64 -48 -16 =0
So its zero.

So x =4 &
x-4=0
but how to get:

(x^2 + 4x +4) =0

Zulfi.
This looks to be basically same problem as in your other recent post.

You divide ##\ x^3 -12x -16 \ ## by ##\ x-4 \ ## using polynomial long division or its short cut, synthetic division.

Homework Helper
Dearly Missed

## Homework Statement

I have following eq:

(x)^3 -12(x) -16=0

How to find x-4 a factor of this eq. This means we have:
(x-4) (x^2 + 4x +4) =0

I don’t know how to get the above.

## Homework Equations

(x)^3 -12(x) -16=0

## The Attempt at a Solution

I cant go beyond that:
(x)^3 -12(x) -16=0
X(x^2 -12-16/x) = 0

You need better notation in order to be able to have a meaningful discussion. So, let ##p(x) = x^3 - 12 x - 16.## IF ##x=4## is a root, then ##p(x)## will be divisible by ##x-4##; see, eg., https://www.purplemath.com/modules/factrthm.htm

Once you have verified that ##p(4) =0## you then know for sure that ##x-4## is a factor of ##p(x)##; that is, ##p(x) = (x-4) q(x)## for some polynomial ##q(x).##
You can find ##q(x)## by the standard algebraic process of (polynomial) long division; see, eg.,
https://en.wikipedia.org/wiki/Polynomial_long_division