CBSE Class 10 Mathematics Polynomials MCQs Set 25

Answer: (d) 194

Explanation:
1. x² + 1/x² = 16 − 2 = 14. Then x⁴ + 1/x⁴ = (x² + 1/x²)² − 2 = 196 − 2 = 194.

Teacher's Note:
1. Square twice, subtracting 2 each time.
2. (x² + 1/x²)² = x⁴ + 2 + 1/x⁴.

Answer: (c) −1

Explanation:
1. p(2) = 4 + 6q − 2q = 4 + 4q = 0, so q = −1.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (b) −8

Explanation:
1. p(−2) = −8 + 24 − 8 + k = 8 + k = 0, so k = −8.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (d) 1

Explanation:
1. p(−1) = (−1)¹⁰⁰ + 2(−1)⁹⁹ + k = 1 − 2 + k = k − 1 = 0, so k = 1.

Teacher's Note:
1. An even power of −1 is 1 and an odd power is −1.
2. Apply the factor theorem.

Answer: (c) 12

Explanation:
1. p(−1) = −1 + 6 + 7 = 12.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (b) 0

Explanation:
1. 16x² + 24x + 9 = (4x + 3)². At x = −3/4, 4x + 3 = 0, so the value is 0.

Teacher's Note:
1. Recognise the perfect square (4x + 3)².
2. Substituting directly also gives 16(9/16) − 18 + 9 = 0.

Answer: (a) −3

Explanation:
1. 2x + 1 = 0 gives x = −1/2. The remainder is p(−1/2) = 2(−1/8) + 5(1/4) − 4(−1/2) − 6 = −1/4 + 5/4 + 2 − 6 = −3.

Teacher's Note:
1. Put the divisor equal to zero to find the value of x.
2. Substitute that value into the polynomial.

Answer: (c) 3/2 and −1/3

Explanation:
1. 6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1). Zeroes: 3/2 and −1/3. Sum = 7/6 = −b/a and product = −1/2 = c/a.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (b) 0 and −2

Explanation:
1. 5u² + 10u = 5u(u + 2), so the zeroes are 0 and −2. Sum = −2 = −b/a and product = 0 = c/a.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (a) 12x² − 5x − 2

Explanation:
1. Sum = 2/3 − 1/4 = 5/12 and product = −1/6. The polynomial is k(x² − (5/12)x − 1/6); with k = 12 we get 12x² − 5x − 2.

Teacher's Note:
1. The polynomial is k[x² − (sum)x + product].
2. Choose k to clear the fractions.

Answer: (d) 6 and 2

Explanation:
1. The polynomial is x² − 8x + 12 = (x − 6)(x − 2), so the zeroes are 6 and 2.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (b) x² − 1

Explanation:
1. x² − (0)x + (−1) = x² − 1, with zeroes 1 and −1.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (a) x² − 10x + 18

Explanation:
1. For rational coefficients the other zero is 5 − √7. Sum = 10 and product = 25 − 7 = 18. The polynomial is x² − 10x + 18.

Teacher's Note:
1. Irrational zeroes of this type occur in conjugate pairs when coefficients are rational.
2. Use sum and product.

Answer: (d) x² − x + 1

Explanation:
1. Dividend = divisor × quotient + remainder, so g(x)(x − 2) = x³ − 3x² + x + 2 + 2x − 4 = x³ − 3x² + 3x − 2. Dividing by x − 2 gives g(x) = x² − x + 1.

Teacher's Note:
1. Use the division algorithm p(x) = g(x)q(x) + r(x).
2. Subtract the remainder first.

Answer: (c) (2, 3)

Explanation:
1. Dividing gives quotient x² + 2x + 3 and remainder: (x² + 5)(x² + 2x + 3) = x⁴ + 2x³ + 8x² + 10x + 15, so the remainder is 12x + 18 − 10x − 15 = 2x + 3. Hence p = 2 and q = 3.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (a) 5

Explanation:
1. Sum of the three zeroes = 4. So 1 + (−2) + z = 4, giving z = 5. Check: product of zeroes = 1 × (−2) × 5 = −10 = −d/a.

Teacher's Note:
1. Sum of zeroes of a cubic = −b/a.
2. Verify with the product of zeroes.

Answer: (d) 1/2 and −1

Explanation:
1. (x − 3 − √2)(x − 3 + √2) = x² − 6x + 7 is a factor. Dividing gives 2x² + x − 1 = (2x − 1)(x + 1), with zeroes 1/2 and −1.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (c) 6

Explanation:
1. α + β = 5 and α − β = 1, so α = 3 and β = 2. Then k = αβ = 6.

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.

Answer: (b) It has no real zero

Explanation:
1. For every real x, x⁴ ≥ 0 and 4x² ≥ 0, so f(x) ≥ 6 > 0. The graph never meets the x-axis, so f has no real zero.

Teacher's Note:
1. A sum of non-negative terms plus a positive constant is always positive.
2. No real zero means the graph does not cut the x-axis.

Answer: (d) 4 and 3

Explanation:
1. For ax² + bx + c the sum of zeroes is −b/a = 8/2 = 4 and the product is c/a = 6/2 = 3. (The zeroes are 1 and 3.)

Teacher's Note:
1. Write the given data in symbols, apply the standard result and then check by substitution.
2. Remember: remainder theorem says the remainder on dividing by (x − a) is p(a); (x − a) is a factor exactly when p(a) = 0.