CBSE Class 10 Mathematics Polynomials MCQs Set 24

Answer: (b) (x + 6)(x − 7)

Explanation:
1. Find two numbers with product −42 and sum −1: these are 6 and −7. So x² − x − 42 = x² + 6x − 7x − 42 = (x + 6)(x − 7).

Teacher's Note:
1. Split the middle term into two numbers whose product is the constant term and whose sum is the coefficient of x.
2. Expand your answer to confirm.

Answer: (a) (6 + y)(1 − y)

Explanation:
1. 6 − 5y − y² = 6 + y − 6y − y² = (6 + y) − y(6 + y) = (6 + y)(1 − y).

Teacher's Note:
1. Rearrange the terms, then group them.
2. Check by expanding: (6 + y)(1 − y) = 6 − 6y + y − y².

Answer: (d) (a + 41)(a + 5)

Explanation:
1. We need two numbers with product 205 and sum 46: 41 and 5. So a² + 46a + 205 = (a + 41)(a + 5).

Teacher's Note:
1. Factor the constant 205 = 5 × 41 first.
2. The sum of the factors must equal the coefficient of a.

Answer: (c) (a − b)(b + c)

Explanation:
1. Group the terms: a(b + c) − b(b + c) = (a − b)(b + c).

Teacher's Note:
1. Group terms that share a common factor.
2. Take the common bracket out.

Answer: (a) 4

Explanation:
1. By the remainder theorem the remainder is p(1) = 2 − 5 + 7 = 4.

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) −5

Explanation:
1. The remainder is p(−1) = (−1)⁹ − 5(−1)⁴ + 1 = −1 − 5 + 1 = −5.

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) −37

Explanation:
1. The remainder is p(−2) = 2(−8) − 3(4) + 4(−2) − 1 = −16 − 12 − 8 − 1 = −37.

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) 6x + 5

Explanation:
1. 36x² + 60x + 25 = (6x)² + 2(6x)(5) + 5² = (6x + 5)², so the positive square root is 6x + 5.

Teacher's Note:
1. Check whether the expression is a perfect square of a binomial, using (a + b)² = a² + 2ab + b².
2. The middle term must equal twice the product of the two roots.

Answer: (d) (x² + 4y + 7)(x² + 4y + 14)

Explanation:
1. Put t = x² + 4y. Then the expression is t² + 21t + 98 = (t + 7)(t + 14), since 7 × 14 = 98 and 7 + 14 = 21. Replace t to get (x² + 4y + 7)(x² + 4y + 14).

Teacher's Note:
1. Substitute a single letter for the repeated bracket.
2. Factorise the quadratic in that letter.

Answer: (c) −2

Explanation:
1. Since x − 2 is a factor, p(2) = 0: 2(8) − 6(4) + 5(2) + k = 16 − 24 + 10 + k = 2 + k = 0, so k = −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) 11

Explanation:
1. Since x + 3 is a factor, p(−3) = 0: 3(9) − 3k + 6 = 33 − 3k = 0, so k = 11.

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) 0

Explanation:
1. The remainder is p(1) = 3 − 7 + 7 − 3 + 2 − 2 = 0. So x − 1 is a factor of the polynomial.

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) m = 2, n = −1

Explanation:
1. p(1) = m + 1 − 2 + n = 0 gives m + n = 1. p(−1) = −m + 1 + 2 + n = 0 gives n − m = −3. Adding: 2n = −2, so n = −1 and m = 2.

Teacher's Note:
1. Use the factor theorem for each factor to get two equations.
2. Solve the equations simultaneously.

Answer: (b) −4 and 3/2

Explanation:
1. 2x² + 5x − 12 = 2x² + 8x − 3x − 12 = 2x(x + 4) − 3(x + 4) = (x + 4)(2x − 3). Zeroes: −4 and 3/2. Check: sum = −5/2 = −b/a and product = −6 = c/a.

Teacher's Note:
1. For a quadratic ax² + bx + c, sum of zeroes = −b/a and product = c/a.
2. Always verify with these two relations.

Answer: (a) √2 and −√2

Explanation:
1. x² − 2 = (x − √2)(x + √2), so the zeroes are √2 and −√2. Sum = 0 = −b/a and product = −2 = c/a.

Teacher's Note:
1. Use a² − b² = (a + b)(a − b) with b = √2.
2. Check sum and product against −b/a and c/a.

Answer: (d) 2√3 and 2/√3

Explanation:
1. √3x² − 8x + 4√3 = √3x² − 6x − 2x + 4√3 = √3x(x − 2√3) − 2(x − 2√3) = (x − 2√3)(√3x − 2). Zeroes: 2√3 and 2/√3. Check: sum = 2√3 + 2/√3 = 8/√3 = −b/a, product = 4 = c/a.

Teacher's Note:
1. Split the middle term using the product ac = 12.
2. Verify the sum and product of zeroes.

Answer: (b) x³ − 6x² − x + 30

Explanation:
1. Sum = 6, sum of products = 15 − 6 − 10 = −1, product = −30. Polynomial: x³ − 6x² − x + 30. (Or expand (x − 3)(x − 5)(x + 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: (a) quotient 5x − 3, remainder −5

Explanation:
1. Write the divisor as x² − 2x + 3. Dividing gives 5x³ − 13x² + 21x − 14 = (x² − 2x + 3)(5x − 3) − 5. Check: (x² − 2x + 3)(5x − 3) = 5x³ − 13x² + 21x − 9, and −9 − 5 = −14.

Teacher's Note:
1. Arrange both polynomials in descending powers of x before dividing.
2. Verify with dividend = divisor × quotient + remainder.

Answer: (d) 5

Explanation:
1. 3x − 2 is a factor when x = 2/3 is a zero. p(2/3) = 3(8/27) + 10(4/9) − 14(2/3) + 9 = 8/9 + 40/9 − 84/9 + 81/9 = 5. So subtract 5.

Teacher's Note:
1. Divisibility by 3x − 2 means the value at x = 2/3 must be zero.
2. The remainder is the number to be subtracted.

Answer: (c) 1 and 1/2

Explanation:
1. (x − √3)(x + √3) = x² − 3 is a factor. Dividing, 2x⁴ − 3x³ − 5x² + 9x − 3 = (x² − 3)(2x² − 3x + 1) = (x² − 3)(2x − 1)(x − 1). So the other zeroes are 1/2 and 1.

Teacher's Note:
1. Divide by the quadratic factor made from the known zeroes.
2. Factorise the quotient.

Answer: (a) 9

Explanation:
1. Squaring: x² − 2 + 1/x² = 1/4, so x² + 1/x² = 9/4. Then 4x² + 4/x² = 4 × 9/4 = 9.

Teacher's Note:
1. Use (x − 1/x)² = x² + 1/x² − 2.
2. Multiply by 4 at the end.