Remainder Theorem...3

[nycmathguy]

Section 2.3
Question 48

If zero is at k= 1/5, one factor of the given polynomial is (x-(1/5))=(x - 1/5).

(1/5)(10) = 2

-22 + 2 = -20

(1/5)(-20) = -4

-3 + (-4) = -7

(1/5)(-7) = -7/5

-7/5 + 4 = 13/5

I say the answer is f(x) = 2x^2 - 20x - 7 with
remainder 13/5.

You say?

 

[MathLover1]

48.
10x^3-22x^2-3x +4=0
x=1/5

..1/5.| 10.................. -22..................-3........................ 4

......... ...... (1/5)10=2.......(1/5)(-20)=-4.....(1/5)(-7)=-7/5
_________________________________________
.........10..................-20..................-7.........................13/5

resulting coefficients: 10,-20,-7, 13/5

the quotient is 10x^2-20x-7, and the remainder is 13/5

 

[nycmathguy]

48.
10x^3-22x^2-3x +4=0
x=1/5

..1/5.| 10.................. -22..................-3........................ 4

......... ...... (1/5)10=2.......(1/5)(-20)=-4.....(1/5)(-7)=-7/5
_________________________________________
.........10..................-20..................-7.........................13/5

resulting coefficients: 10,-20,-7, 13/5

the quotient is 10x^2-20x-7, and the remainder is 13/5

I did not include 10 in the list of coefficients. I will try again. For the most part, not too bad.