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Class – X – NCERT – Maths, , Polynomials, , Solution:, (i), , Since the graph of 𝒑(𝑥) does not cut the X-axis at all. Therefore, the, number of zeroes is 𝟎., , (ii), , As the graph of 𝒑(𝑥) intersects the X-axis at only 𝟏 point. Therefore, the, number of zeroes is 𝟏., , (iii), , Since the graph of 𝒑(𝑥) intersects the X-axis at 𝟑 points. Hence, the, number of zeroes is 3., , (iv), , As the graph of 𝒑(𝑥) intersects the X-axis at 𝟐 points. So, the number of, zeroes is 𝟐., , (v), , Since the graph of 𝒑(𝑥) intersects the X-axis at 𝟒 points. Therefore, the, number of zeroes is 𝟒., , (vi), , As the graph of 𝒑(𝑥) intersects the X-axis at 𝟑 points. So, the number of, zeroes is 𝟑.,   , , CBSE NCERT Solutions for Class 10 Science Chapter 2 – Ex, 2.2, 1., , Find the zeroes of the following quadratic polynomials and verify the relationship, between the zeroes and the coefficients., (i), , 𝑥 2 − 2𝑥 − 8, , (ii), , 4𝑠 2 − 4𝑠 + 1, , (iii), , 6𝑥 2 − 3 − 7𝑥, , (iv), , 4𝑢2 + 8𝑢, Page - 3

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Class – X – NCERT – Maths, , Polynomials, , ∴ 𝑔(𝑥) = 𝑥 2 − 𝑥 + 1, , 6., , Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the, division algorithm and, (i), , deg p(x) = deg q(x), , (ii), , deg q(x) = deg r(x), , (iii), , deg r(x) = 0, , Solution:, According to the division algorithm, if p(x) and g(x) are two polynomials with, g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that, p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x)., (i), , Degree of quotient will be equal to degree of dividend when divisor is, constant., Let us consider the division of 2x 2 + 2x − 16 by 2., Here, p(x) = 2x 2 + 2x − 16 and g(x) = 2, q(x) = x 2 + x − 8 and r(x) = 0, Clearly, the degree of p(x) and q(x) is the same which is 2., Verification:, p(x) = g(x) × q(x) + r(x), Page - 16

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Class – X – NCERT – Maths, , Polynomials, , 2x 2 + 2x − 16 = 2(x 2 + x − 8) + 0, = 2x 2 + 2x − 16, Thus, the division algorithm is satisfied., (ii), , Let us consider the division of 4x + 3 by x + 2., Here, p(x) = 4x + 3 and g(x) = x + 2, q(x) = 4 and r(x) = −5, Here, degree of q(x) and r(x) is the same which is 0., Verification:, p(x) = g(x) × q(x) + r(x), 4x + 3 = (x + 2) × 4 + (−5), 4x + 3 = 4x + 3, Thus, the division algorithm is satisfied., , (iii), , Degree of remainder will be 0 when remainder obtained on division is a, constant., Let us consider the division of 4x + 3 by x + 2., Here, p(x) = 4x + 3 and g(x) = x + 2, q(x) = 4 and r(x) = −5, Here, we get remainder as a constant. Therefore, the degree of r(x) is 0., Verification:, p(x) = g(x) × q(x) + r(x), 4x + 3 = (x + 2) × 4 + (−5), 4x + 3 = 4x + 3, Thus, the division algorithm is satisfied., ♦ ♦ ♦, , Page - 17