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Find the factors of an equation

  • Thread starter zak100
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  • #1
zak100
Gold Member
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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

  • #2
berkeman
Mentor
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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?
 
  • #3
zak100
Gold Member
435
10
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.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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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.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
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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
 

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