If the polynomial 3x^3-4x^2-17x+17+k is exactly divisible by 3x-1 find k.
Answers
Answered by
2
3x-1=0
x=1/3
As the given polynomial is exactly divisible by 3x-1,
f(x)=3x^3-4x^2-17x+17+k
f(1/3)=3(1/3)^3-4(1/3)^2-17*1/3+17+k
0=(1/9)-(4/9)-(17/3)+17+k
0= (-3/9)-(17/3)+17+k
0= -(54/9)+17+k
k= (54/9)-(153/9)
k= -99/9
k= -11
x=1/3
As the given polynomial is exactly divisible by 3x-1,
f(x)=3x^3-4x^2-17x+17+k
f(1/3)=3(1/3)^3-4(1/3)^2-17*1/3+17+k
0=(1/9)-(4/9)-(17/3)+17+k
0= (-3/9)-(17/3)+17+k
0= -(54/9)+17+k
k= (54/9)-(153/9)
k= -99/9
k= -11
Similar questions