Question

p(x) = x³ - 9x² + 20x - 12, d(x) = x - 6By using factor theorem, find the other factor of the above polynomial p(x), where d(x) is a given factor.

Answer 1

x-6)x^3-9x^2+20x-12(x^2-3x+2
x^3-6x^2
___-__+_______
-3x^2+20x
-3x^2+18x
____+____-______
2x-12
2x-12
___-__+_____
0
x^2-3x+2=0
x^2-2x-x+2=0
x(x-2)-1(x-2)=0
(x-2)(x-1)=0
(x-2) and (x-1) are the other two factors

Answer 2

Hi ,

********************************************
Factor Theorem :

Let p(x) be a polynomial of degree

one or more than 1 and a real number.

Then ,

i ) x - a , will be a factor of p(x) if p(a)=0

*************************************************

Here ,

p(x) = x³ - 9x² + 20x - 12 ,

d(x) = x - 6

p(6) = 6³ - 9×6² + 20×6 - 12

= 216 - 324 + 120 - 12

= 336 - 336

= 0

Therefore ,

( x - 6 ) is a factor of p(x).

2 )

x - 6 ) x³ - 9x² + 20x - 12 ( x²- 3x+2
******* x³ - 6x²
_________________
*********** -3x² + 20x
*********** -3x² + 18x
_________________
******************* 2x - 12
******************* 2x - 12
__________________
Remainder ( 0 )

p(x) = ( x - 6 )( x² - 3x + 2 )

= ( x - 6 )[ x² - x - 2x + 2 ]

= ( x - 6 ) [ x( x - 1 ) - 2( x - 1 ) ]

= ( x - 6 )( x - 1 )( x - 2 )

I hope this helps you.

: )