3h. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.8)
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Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Factorization using synthetic division
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The document contains solutions to 26 math problems involving multiplying polynomials. The problems involve multiplying terms with variables like x, y, a, b. The solutions show the distribution of terms and combining like terms. Some problems then verify the solutions by plugging in values for the variables. The document provides worked out solutions to multiplying polynomials of varying complexities.
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- 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
- 2. STD IX CHAPTER 3 – ALGEBRA Ex – 3.8
- 10. (i) x3 – 3x² – 10x + 24 Solution: p(x) – x3 – 3x² – 10x + 24 p(1) = 13 – 3(1)² – 10(1) + 24 = 1 – 3 – 10 + 24 = 25 – 13 ≠ 0 x – 1 is not a factor p(-1) = (-1)3 – 3(-1)² – 10(-1) + 24 = – 1 – 3(1) + 10 + 24 = -1 – 3 + 10 + 24 = 34 – 4 = 30 ≠ 0 x + 1 is not a factor
- 11. p(2) = 23 – 3(2)² – 10(2) + 24 = 8 – 3(4) – 20 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0 ∴ x – 2 is a factor x² – x – 12 = x² – 4x + 3x – 12 = x(x – 4) + 3 (x – 4) = (x – 4) (x + 3) ∴ The factors of x3 – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)
- 12. (ii) 2x3 – 3x² – 3x + 2 Solution: p(x) = 2x3 – 3x² – 3x + 2 P(1) = 2(1)3 – 3(1)² – 3(1) + 2 = 2 – 3 – 3 + 2 = 2 – 6 = -4 ≠ 0 x – 1 is not a factor P(-1) = 2(-1)3 – 3(-1)² – 3(-1) + 2 = -2 – 3 + 3 + 2 = 5 – 5 = 0 ∴ x + 1 is a factor
- 13. 2x² – 5x + 2 = 2x² – 4x – x + 2 = 2x(x – 2) – 1 (x – 2) = (x – 2) (2x – 1) ∴The factors of 2x3 – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)
- 14. (iii) – 7x + 3 + 4x3 Solution: p(x) = – 7x + 3 + 4x3 = 4x3 – 7x + 3 P(1) = 4(1)3 – 7(1) + 3 4 – 7 + 3 = 7 – 7 = 0 ∴ x – 1 is a factor 4x² + 4x – 3 = 4x² + 6x – 2x – 3 = 2x(2x + 3) – 1 (2x + 3) = (2x + 3) (2x – 1) ∴ The factors of – 7x + 3 + 4x3 = (x – 1) (2x + 3)
- 15. (iv) x3 + x² – 14x – 24 Solution: p(x) = x3 + x² – 14x – 24 p(1) = (1)3 + (1)2 – 14 (1) – 24 = 1 + 1 – 14 – 24 = -36 ≠ 0 x + 1 is not a factor. p(-1) = (-1)3 + (-1)² – 14(-1) – 24 = -1 + 1 + 14 – 24 = 15 – 25 ≠ 0 x – 1 is not a factor.
- 16. = x(x – 4) + 3 (x – 4) = (x – 4) (x + 3) This (x + 2) (x + 3) (x – 4) are the factors. x3 + x2 – 14x – 24 = (x + 2) (x + 3) (x – 4)
- 17. (v) x3 – 7x + 6 Solution: p(x) = x3 – 7x + 6 P( 1) = 13 – 7(1) + 6 = 1 – 7 + 6 = 7 – 7 = 0 ∴ x – 1 is a factor
- 18. (vi) x3 – 10x² – x + 10 p(x) = x3 – 10x2 – x + 10 = 1 – 10 – 1 + 10 = 11 – 11 = 0 ∴ x – 1 is a factor x2 – 9x – 10 = x2 – 10x + x – 10 = x(x – 10) + 1 (x – 10) = (x – 10) (x + 1) This (x – 1) (x + 1) (x – 10) are the factors. ∴ x3 – 10x2 – x + 10 = (x – 1) (x – 10) (x + 1)