Given, p(x)=5x−π
On substituting x=45 in p(x), we get,
p(45)=5(45)−π=4−π
Since p(x)≠0 at x=45 ,
therefore, x=45 is not a zero of polynomial p(x)=5x−π
That is, option B is correct.
Verify whether the following values indicated against them are zeroes of the polynomial. (i)p(x)=5x−π,x=45 (ii)p(x)=x2−1,x=1,−1 [4 MARKS]