Mathematics 10 — Exercise 5.2 · Q1 (v)

Full question

Simplify the following sum or difference of algebraic fractions.

1x+1−2x+2+-2×x-3x2+3×x+2

Step-by-step solution

  1. Fraction 1: keep its denominator nonzero and factor that denominator.

    x+1≠0
  2. Fraction 2: keep its denominator nonzero and factor that denominator.

    x+2≠0
  3. Fraction 3: keep its denominator nonzero and factor that denominator.

    x2+3×x+2≠0
  4. Expand any powers or inner brackets before factoring.

    x2+3×x+2=x2+3×x+2
  5. Split the middle coefficient: these two numbers add to b and multiply to a·c.

    (2+1=3,2×1=2)
  6. Rewrite the middle term as those two terms.

    x2+2×x+x+2
  7. Group two terms at a time and take each common factor.

    x×(x+2)+x+2
  8. Factor out the common bracket.

    (x+1)×(x+2)
  9. The complete factored form is:

    (x+1)×(x+2)
  10. Build the least common denominator from every distinct denominator factor at its largest required power.

    (x+1)×(x+2)
  11. For fraction 1, divide the common denominator by its own denominator.

    x2+3×x+2x+1=x+2
  12. Multiply this numerator by the same factor to keep the fraction unchanged.

    x+2(x+1)×(x+2)
  13. For fraction 2, divide the common denominator by its own denominator.

    x2+3×x+2x+2=x+1
  14. Multiply this numerator by the same factor to keep the fraction unchanged.

    -2×(x+1)(x+1)×(x+2)
  15. For fraction 3, divide the common denominator by its own denominator.

    x2+3×x+2x2+3×x+2=1
  16. Multiply this numerator by the same factor to keep the fraction unchanged.

    1×(-2×x-3)(x+1)×(x+2)
  17. The denominators now match. Put all signed numerators over one common denominator.

    x+2−2×(x+1)+1×(-2×x-3)(x+1)×(x+2)
  18. Expand numerator product 1, keeping minus signs on the whole bracket.

    x+2=x+2
  19. Expand numerator product 2, keeping minus signs on the whole bracket.

    -2×(x+1)=-2×x-2
  20. Expand numerator product 3, keeping minus signs on the whole bracket.

    1×(-2×x-3)=-2×x-3
  21. Collect like powers and constants in the combined numerator.

    x+2−(2×x+2)−(2×x+3)=-3×x-3
  22. The original denominator must be nonzero. Keep this restriction even if a factor is later cancelled.

    x2+3×x+2≠0
  23. Factor the numerator.

  24. Expand any powers or inner brackets before factoring.

    -3×x-3=-3×x-3
  25. Take the common factor out of every term. Show each remaining term after division.

    -3×x-3=3×(-x-1)
  26. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    -3×x-3=-3×(x+1)
  27. The complete factored form is:

    -3×(x+1)
  28. Factor the denominator.

  29. Expand any powers or inner brackets before factoring.

    x2+3×x+2=x2+3×x+2
  30. Split the middle coefficient: these two numbers add to b and multiply to a·c.

    (2+1=3,2×1=2)
  31. Rewrite the middle term as those two terms.

    x2+2×x+x+2
  32. Group two terms at a time and take each common factor.

    x×(x+2)+x+2
  33. Factor out the common bracket.

    (x+1)×(x+2)
  34. The complete factored form is:

    (x+1)×(x+2)
  35. Write the factored numerator over the factored denominator.

    -3×(x+1)(x+1)×(x+2)
  36. This is the common factor of numerator and denominator.

    x+1
  37. Divide numerator and denominator by that same nonzero factor.

    -3×x-3x+1x2+3×x+2x+1
  38. Simplify the two divisions separately.

    -3x+2
  39. Write the fraction in lowest form. Only factors, never added terms, have been cancelled.

    -3x+2

Answer

The final answer is shown below.

-3x+2