Mathematics 10 — Exercise Review · Q3 (i)

Full question

Simplify the following sum or difference of algebraic fractions.

5x2+x-6−12×x2-7×x+6

Step-by-step solution

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

    x2+x-6≠0
  2. Expand any powers or inner brackets before factoring.

    x2+x-6=x2+x-6
  3. Split the middle coefficient: these two numbers add to b and multiply to a·c.

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

    x2+3×x−2×x−6
  5. Group two terms at a time and take each common factor.

    x×(x+3)−2×(x+3)
  6. Factor out the common bracket.

    (x-2)×(x+3)
  7. The complete factored form is:

    (x-2)×(x+3)
  8. Fraction 2: keep its denominator nonzero and factor that denominator.

    2×x2-7×x+6≠0
  9. Expand any powers or inner brackets before factoring.

    2×x2-7×x+6=2×x2-7×x+6
  10. Split the middle coefficient: these two numbers add to b and multiply to a·c.

    (−3−4=-7,(-3)×(-4)=12)
  11. Rewrite the middle term as those two terms.

    2×x2−3×x−4×x+6
  12. Group two terms at a time and take each common factor.

    x×(2×x-3)−2×(2×x-3)
  13. Factor out the common bracket.

    (x-2)×(2×x-3)
  14. The complete factored form is:

    (x-2)×(2×x-3)
  15. Build the least common denominator from every distinct denominator factor at its largest required power.

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

    2×x3-x2-15×x+18x2+x-6=2×x-3
  17. Multiply this numerator by the same factor to keep the fraction unchanged.

    5×(2×x-3)(x-2)×(x+3)×(2×x-3)
  18. For fraction 2, divide the common denominator by its own denominator.

    2×x3-x2-15×x+182×x2-7×x+6=x+3
  19. Multiply this numerator by the same factor to keep the fraction unchanged.

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

    5×(2×x-3)−(x+3)(x-2)×(x+3)×(2×x-3)
  21. Expand numerator product 1, keeping minus signs on the whole bracket.

    5×(2×x-3)=10×x-15
  22. Expand numerator product 2, keeping minus signs on the whole bracket.

    -(x+3)=-x-3
  23. Collect like powers and constants in the combined numerator.

    10×x-15−(x+3)=9×x-18
  24. The original denominator must be nonzero. Keep this restriction even if a factor is later cancelled.

    2×x3-x2-15×x+18≠0
  25. Factor the numerator.

  26. Expand any powers or inner brackets before factoring.

    9×x-18=9×x-18
  27. Take the common factor out of every term. Show each remaining term after division.

    9×x-18=9×(x-2)
  28. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    9×x-18=9×(x-2)
  29. The complete factored form is:

    9×(x-2)
  30. Factor the denominator.

  31. Expand any powers or inner brackets before factoring.

    2×x3-x2-15×x+18=2×x3-x2-15×x+18
  32. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    2×x3-x2-15×x+18=(x-2)×(x+3)×(2×x-3)
  33. The complete factored form is:

    (x-2)×(x+3)×(2×x-3)
  34. Write the factored numerator over the factored denominator.

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

    x-2
  36. Divide numerator and denominator by that same nonzero factor.

    9×x-18x-22×x3-x2-15×x+18x-2
  37. Simplify the two divisions separately.

    92×x2+3×x-9
  38. Write the fraction in lowest form. Only factors, never added terms, have been cancelled.

    9(x+3)×(2×x-3)

Answer

The final answer is shown below.

9(x+3)×(2×x-3)