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Kerala SSLC (Class 10) Real Numbers and Polynomials — practice questions — page 4

Explore 102 free Real Numbers and Polynomials questions in Kerala SSLC (Class 10) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Real Numbers and Polynomials questions and exercises with answers

Showing questions 91–102 of 102.

The solutions of the equation x² + 2x − 15 = 0 are:Read the answer and worked solution →If p(x) is a polynomial and p(a) = 0 for some number a, then which first-degree polynomial is a factor of p(x)?Read the answer and worked solution →Which first-degree polynomial is a factor of x⁵ − a⁵ for every value of a?Read the answer and worked solution →Using the quadratic formula, the solutions of x² − 30x + 221 = 0 are:Read the answer and worked solution →The polynomial x² + 1 has no first-degree factors over the real numbers because:Read the answer and worked solution →In the polynomial x² + kx + 6, the value of k that makes x − 1 a factor is:Read the answer and worked solution →The factorization of 2x² − 7x + 6 into first-degree polynomials is:Read the answer and worked solution →Which of the following second-degree polynomials cannot be factored into first-degree real polynomials?Read the answer and worked solution →Given any polynomial p(x) and any number a, which polynomial is always a factor of p(x) − p(a)?Read the answer and worked solution →Match each second-degree polynomial to its first-degree factorization.Read the answer and worked solution →Tap TRUE or FALSE for each polynomial / equation statement.Read the answer and worked solution →Fill the polynomial-factor blanks so each equation is true.Read the answer and worked solution →

Worked practice questions

The solutions of the equation x² + 2x − 15 = 0 are:

  1. 3 and 5
  2. 3 and −5
  3. −3 and 5
  4. −3 and −5
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Correct answer: C. −3 and 5

1. Write x² + 2x − 15 = (x − a)(x − b), requiring a + b = −2 and ab = −15. 2. Factor pairs of 15: 3 × 5. One must be negative since the product is negative. 3. Trying a = −3, b = 5: a + b = 2, ab = −15 — sum is wrong. 4. Trying a = 3, b = −5: a + b = −2, ab = −15 — both match. 5. So x² + 2x − 15 = (x − 3)(x + 5) and the solutions are 3 and −5. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 9-11 (pp 227-246, 2019 ed.)._

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If p(x) is a polynomial and p(a) = 0 for some number a, then which first-degree polynomial is a factor of p(x)?

  1. x + a
  2. a − x
  3. ax + 1
  4. x − a
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Correct answer: D. x − a

1. The factor theorem links roots of p(x) to first-degree factors. 2. If p(a) = 0, then x − a is a factor of p(x). 3. This is the unit's Factor Theorem statement. 4. Conversely, if x − a is a factor of p(x), then p(a) = 0. 5. So the correct factor is x − a, not x + a or a − x. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 9-11 (pp 227-246, 2019 ed.)._

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Which first-degree polynomial is a factor of x⁵ − a⁵ for every value of a?

  1. x − a
  2. x² − a²
  3. x + a
  4. x − a²
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Correct answer: A. x − a

1. The unit's 'Power difference' box states that x − a is a factor of xⁿ − aⁿ for every natural number n. 2. For n = 5 the result xⁿ − aⁿ = x⁵ − a⁵. 3. So x − a is the first-degree factor. 4. The general formula gives x⁵ − a⁵ = (x − a)(x⁴ + ax³ + a²x² + a³x + a⁴). 5. The other options are higher-degree or wrong-sign factors. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 9-11 (pp 227-246, 2019 ed.)._

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