Verify that:
(i) 2 is a zero of the polynomial p(x) = (x $-$ 2).
(ii) 2 and 9 are zeroes of the polynomial p(x) = (x $-$ 2) (x $-$ 9).
(iii) 4 and $-$ 3 are zeroes of the polynomial p(x) = x² $-$ x $-$ 12.

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Solution

$(i) We have p (x) = x - 2 \Rightarrow p (2) = 2 - 2 \Rightarrow p (2) = 0 ∴ 2 is a zero of the given polynomial .$

$(ii) We have p (x) = (x - 2) (x - 9) \Rightarrow p (2) = (2 - 2) (2 - 9) \Rightarrow p (2) = 0 \times (- 7) \Rightarrow p (2) = 0 and p (9) = (9 - 2) (9 - 9) \Rightarrow p (9) = 7 \times 0 \Rightarrow p (9) = 0 ∴ 2 and 9 are zero e s of the given polynomial .$

$(iii) We have p (x) = x^{2} - x - 12 \Rightarrow p (4) = 4^{2} - 4 - 12 \Rightarrow p (4) = 16 - 4 - 12 \Rightarrow p (4) = 0 and p (- 3) = {(- 3)}^{2} - (- 3) - 12 \Rightarrow p (- 3) = 9 + 3 - 12 \Rightarrow p (- 3) = 0 ∴ 2 and - 3 are zeroes of the given polynomial .$

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Verify that: (i) 2 is a zero of the polynomial p(x) = (x - 2). (ii) 2 and 9 are zeroes of the polynomial p(x) = (x - 2) (x - 9). (iii) 4 and - 3 are zeroes of the polynomial p(x) = x2 - x - 12.

Verify that:
(i) 2 is a zero of the polynomial p(x) = (x $-$ 2).
(ii) 2 and 9 are zeroes of the polynomial p(x) = (x $-$ 2) (x $-$ 9).
(iii) 4 and $-$ 3 are zeroes of the polynomial p(x) = x² $-$ x $-$ 12.