Question

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = - \(\frac{1}{3}\)

(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)

(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) = lx + m, x = – \(\frac{m}{l}\)

(vii) p(x) = 3x ² – 1, x = -\(\frac{1 }{ √3} ,\frac{2}{ √3} \)

(viii) p(x) = 2x + 1, x = \(\frac{1}{2}\)

Answer 1

(i) If x = -\(\frac{1}{3} \) is a zero of given polynomial p(x) = 3x + 1, then p(-\(\frac{1}{3} \)) should be 0. 

Here, p (-\(\frac{1}{3} \)) = 3 (-\(\frac{1}{3} \)) + 1 = -1 + 1 = 0. 

Therefore, x = -\(\frac{1}{3} \) is a zero of the given polynomial.

(ii) If x = \(\frac{4}{5} \) is a zero of polynomial p(x) = 5x - π, then p(4/5) should be 0. 

Here, p(\(\frac{4}{5} \)) = 5 (\(\frac{4}{5} \)) - π = 4 - π As p (4/5) ≠ 0, 

therefore, x = \(\frac{4}{5} \) is not a zero of the given polynomial.

(iii) If x = 1 and y = -1 are zeroes of polynomial p(x) = x2 - 1, then p(1) and p(-1) should be 0. 

Here, p(1) = (1)2 - 1 = 0, and p(-1) = (-1)2 - 1 = 0 

Hence, x = 1 and −1 are zeroes of the given polynomial.

(iv)If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0. 

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0 

Therefore, x = −1 and x = 2 are zeroes of the given polynomial.

(v) If x = 0 is a zero of polynomial p(x) = x2 , then p(0) should be zero. 

Here, p(0) = (0)2 = 0. 

Hence, x = 0 is a zero of the given polynomial.

(vi) If x = -\(\frac{m }{ l} \)is a zero of polynomial p(x) = lx + m, then should be 0. 

Here, p(-\(\frac{m }{ l} \) )= (-\(\frac{m }{ l} \)) + m = -m + m = 0. 

Therefore, x = -\(\frac{m }{ l} \)l is a zero of the given polynomial.

(vii) If x = -\(\frac{1 }{ √3} \)and x = \(\frac{2}{ √3} \) are zeroes of polynomial 

p(x) = 3x2 -1 , then p(-\(\frac{m}{l}\))

p(-\(\frac{1}{√3}\)) and p (\(\frac{2}{√3}\)) should be 0. 

Here, p(-\(\frac{1}{√3}\)) = 3 (-\(\frac{1}{√3}\))2 -1 = 3 (\(\frac{1}{3}\))-1 = 1-1 = 0, and 

p(2/√3) = 3 (\(\frac{2}{√3}\))3 - 1 = 3 (\(\frac{2}{√3}\))2 -1 = 3(\(\frac{4}{3}\)) - 1 = 4 - 1 = 3. 

Hence, x = -\(\frac{1}{√3}\) is a zero of the given polynomial. 

However, x = 2√3 is not a zero of the given polynomial.

(viii) If x = \(\frac{1}{2}\) is a zero of polynomial p(x) = 2x + 1, then p(1/2) should be 0. 

Here, p (\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) + 1 = 1 +1 = 2. As p ≠ 0, 

Therefore, x = \(\frac{1}{2}\) 2 is not a zero of given polynomial .