Verify whether the following are zeroes of the polynomial, indicated against them. (i) `p(x)=3x+1, x=-1/3` (ii) `p(x)=5x-pi, x=4/5` (iii) `p(x)=x^2-1, x=1,-1` (iv) `p(x)=(x+1)(x+2), x=-1,2` (v) `p(x)=x^2, x=0` (vi) `p(x)=l x+m , x=-m/l` (vii)

Explanation

(i)`p(x)=3x+1,x=-1/3`
`=>p(-1/3)=3(-1/3)+1`
`=>p(-1/3)=0`
Therefore ,`x=-1/3` is zero of p(x)
(ii)`p(x)=5x-pi,x=4/5`
`p(4/5)=5(4/5)-pi=4-pi`Now `pi=22/7`not equal to zero.
So `x=4/5` is not zero of p(x)
(iii)`p(x)=x^2-1,x=1,-1`
`x=1,=>p(1)=(1)^2-1=0`
`x=-1,=>p(-1)=(-1)^2-1=0`
So `x=1,-1` is zeroes of p(x)
(iv) `p(x)=(x+1)(x+2),x=-1,2`
`=>p(-1)=(-1+1)(1+2)=0`,
`=>p(2)=(2+1)(2+2)=12`,So x=-1 is zero of p(x)
but `x=2` is not zero of p(x)
(v) `p(x)=x^2, x=0`,`p(0)=(0)^2=0`
so`x=0` is zero of p(x)
(vi) `p(x)=l x+m , x=-m/l`
`=>p(m/l)=l(-m/l)+m=0`,
So, `x=m/l` is zero of p(x).