If x-2y =0, then in this case x and y vary in what way?
Answers
Step-by-step explanation:
, -1, 2 are the zeroes of the polynomial x
3
−2x
2
−x+2
Step-by-step explanation:
Given the cubic polynomial
P(x)=x^3-2x^2-x+2P(x)=x
3
−2x
2
−x+2
we have to find the zeroes of the polynomial
P(x)=x^3-2x^2-x+2P(x)=x
3
−2x
2
−x+2
By hit and trial method put x=1
P(1)=1^3-2(1)^2-1+2P(1)=1
3
−2(1)
2
−1+2
p(1)1-2-1+2=0p(1)1−2−1+2=0
which implies (x-1) is the factor of given polynomial or 1 is the zero of polynomial.
By synthetic division shown in attachment
x^3-2x^2-x+2=(x-1)(x^2-x-2)x
3
−2x
2
−x+2=(x−1)(x
2
−x−2)
(x-1)(x^2-2x+x-2(x−1)(x
2
−2x+x−2
(x-1)(x(x-2)+1(x-2))(x−1)(x(x−2)+1(x−2))
(x-1)(x+1)(x-2)(x−1)(x+1)(x−2)
Hence, the zeroes of polynomial are
x-1=0 ⇒ x=1
x+1=0 ⇒ x=-1
x-2=0 ⇒ x=2
Answer:
The general form of a direct variation equation is
y=kx
, with k being the constant of variation.
−x+2y=0
can be transformed to fit the correct form:
−x+x+2y=0+x
2y=x2y2=x2
y=12x
Therefore, it is a direct variation equation and
k=12
.