Factor Theorem...3

[nycmathguy]

Section 2.3
Question 58

If zero is at x = 2/3 one factor of the given polynomial is (x - 2/3).

(2/3)(48) = 16

-80 + 16 = -64

(2/3)(-64) = (-128/3)

41 - (-128/3) = (-256/9)

-6 - (512/27) = -(674/27)

I say the answer is 48x^2 - 64x - (256/9) with remainder -(674/27).

You say?

 

[MathLover1]

58.
48x^3-80x^2+41x -6=0
x=2/3

if x=2/3 is a root, then (x-2/3) is a factor and reminder is 0

..2/3.| 48.................. -80..................41.................. -6

......... .... (2/3)48=32.......(2/3)(-48)=-32.....(2/3)9=6
_________________________________________
.........48..................-48..................9....................0

resulting coefficients: 48, −48, 9, 0

the quotient is 48x^2-48x+9, and the remainder is 0

remember:
first coefficients does not change: 48
second coefficient will be :

-80+ (2/3)(48)=-80 + 32 =-48
third coefficient will be :

41+(2/3)(-48)=41-32=9
and reminder
-6+(2/3)9=-6+6=0

 

[nycmathguy]

58.
48x^3-80x^2+41x -6=0
x=2/3

if x=2/3 is a root, then (x-2/3) is a factor and reminder is 0

..2/3.| 48.................. -80..................41.................. -6

......... .... (2/3)48=32.......(2/3)(-48)=-32.....(2/3)9=6
_________________________________________
.........48..................-48..................9....................0

resulting coefficients: 48, −48, 9, 0

the quotient is 48x^2-48x+9, and the remainder is 0

remember:
first coefficients does not change: 48
second coefficient will be :

-80+ (2/3)(48)=-80 + 32 =-48
third coefficient will be :

41+(2/3)(-48)=41-32=9
and reminder
-6+(2/3)9=-6+6=0

I did it wrong. I gotta try again.