The number of polynomials having zeroes as –2 and 5 is(A) 1 (B) 2(C)

1.	The number of polynomials having zeroes as –2 and 5 is(A) 1 (B) 2(C) 3 (D) more than 3
Answer» (D) more than 3 Explanation: According to the question, The zeroes of the polynomials = -2 and 5 We know that the polynomial is of the form, p(x) = ax² + bx + c. Sum of the zeroes = – (coefficient of x) ÷ coefficient of x² i.e. Sum of the zeroes = – b/a – 2 + 5 = – b/a 3 = – b/a b = – 3 and a = 1 Product of the zeroes = constant term ÷ coefficient of x² i.e. Product of zeroes = c/a (- 2)5 = c/a – 10 = c Substituting the values of a, b and c in the polynomial p(x) = ax² + bx + c. We get, x² – 3x – 10 Therefore, we can conclude that x can take any value. Hence, option (D) is the correct answer.

The number of polynomials having zeroes as –2 and 5 is(A) 1 (B) 2(C) 3 (D) more than 3

Answer»

(D) more than 3

Explanation:

According to the question,

The zeroes of the polynomials = -2 and 5

We know that the polynomial is of the form,

p(x) = ax² + bx + c.

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x² i.e.

Sum of the zeroes = – b/a

– 2 + 5 = – b/a

3 = – b/a

b = – 3 and a = 1

Product of the zeroes = constant term ÷ coefficient of x² i.e.

Product of zeroes = c/a

(- 2)5 = c/a

– 10 = c

Substituting the values of a, b and c in the polynomial p(x) = ax² + bx + c.

We get, x² – 3x – 10

Therefore, we can conclude that x can take any value.

Hence, option (D) is the correct answer.