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) =0I don’t know how to get the above.

Homework Equations

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

The Attempt at a Solution

I can't go beyond that:
(x)^3 -12(x) -16=0
X(x^2 -12-16/x) = 0

 

[berkeman]

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

Please guide me.

Zulfi.

 

[SammyS]

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

Please guide me.

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.

 

[Ray Vickson]

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) =0I don’t know how to get the above.

Homework Equations

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

The Attempt at a Solution

I can't 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