Now, p(−1√3)=−3(−1√3)2−1=−1−1=−2 and
p(2√3)=−3(2√3)2−1=−12−1=−13
Clearly, p(−1√3)≠0 and p(2√3)≠0
∴x=−1√3,2√3 are not zeroes of given polynomial.
Verify whether the following are zeros of the polynomial, indicated against them.
1.. P(x) = 3x + 1, x=1/3
2.. p(y) = (x+1)(x+2), x= -2
explain how to solve these questions
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) (ii)
(iii) p(x) = x2 − 1, x = 1, − 1 (iv) p(x) = (x + 1) (x − 2), x = − 1, 2
(v) p(x) = x2 , x = 0 (vi) p(x) = lx + m
(vii) (viii)