Mathematics 10 — Exercise 5.2 · Q1 (ix)

Full question

Simplify the following sum or difference of algebraic fractions.

Step-by-step solution

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

   $$x^2-x-12\ne 0$$

2. Expand any powers or inner brackets before factoring.

   $$x^2-x-12=x^2-x-12$$

3. Split the middle coefficient: these two numbers add to b and multiply to a·c.

   $$(3-4=-1,3\times \left(-4\right)=-12)$$

4. Rewrite the middle term as those two terms.

   $$x^2+3\times x-4\times x-12$$

5. Group two terms at a time and take each common factor.

   $$x\times \left(x+3\right)-4\times \left(x+3\right)$$

6. Factor out the common bracket.

   $$\left(x-4\right)\times \left(x+3\right)$$

7. The complete factored form is:

   $$\left(x-4\right)\times \left(x+3\right)$$

8. Fraction 2: keep its denominator nonzero and factor that denominator.

   $$x^2+6\times x+9\ne 0$$

9. Expand any powers or inner brackets before factoring.

   $$x^2+6\times x+9=x^2+6\times x+9$$

10. Split the middle coefficient: these two numbers add to b and multiply to a·c.

    $$(3+3=6,3\times 3=9)$$

11. Rewrite the middle term as those two terms.

    $$x^2+3\times x+3\times x+9$$

12. Group two terms at a time and take each common factor.

    $$x\times \left(x+3\right)+3\times \left(x+3\right)$$

13. Factor out the common bracket.

    $$\left(x+3\right)\left(x+3\right)$$

14. The complete factored form is:

    $${\left(x+3\right)}^2$$

15. Build the least common denominator from every distinct denominator factor at its largest required power.

    $$\left(x-4\right)\times {\left(x+3\right)}^2$$

16. For fraction 1, divide the common denominator by its own denominator.

    $$\frac{x^3+2\times x^2-15\times x-36}{x^2-x-12}=x+3$$

17. Multiply this numerator by the same factor to keep the fraction unchanged.

    $$\frac{\left(x+2\right)\times \left(x+3\right)}{\left(x-4\right)\times {\left(x+3\right)}^2}$$

18. For fraction 2, divide the common denominator by its own denominator.

    $$\frac{x^3+2\times x^2-15\times x-36}{x^2+6\times x+9}=x-4$$

19. Multiply this numerator by the same factor to keep the fraction unchanged.

    $$\frac{\left(-x\right)\times \left(x-4\right)}{\left(x-4\right)\times {\left(x+3\right)}^2}$$

20. The denominators now match. Put all signed numerators over one common denominator.

    $$\frac{\left(x+2\right)\times \left(x+3\right)+\left(-x\right)\times \left(x-4\right)}{\left(x-4\right)\times {\left(x+3\right)}^2}$$

21. Expand numerator product 1, keeping minus signs on the whole bracket.

    $$\left(x+2\right)\times \left(x+3\right)=x^2+5\times x+6$$

22. Expand numerator product 2, keeping minus signs on the whole bracket.

    $$\left(-x\right)\times \left(x-4\right)=-x^2+4\times x$$

23. Collect like powers and constants in the combined numerator.

    $$x^2+5\times x+6-(x^2-4\times x)=9\times x+6$$

24. The original denominator must be nonzero. Keep this restriction even if a factor is later cancelled.

    $$x^3+2\times x^2-15\times x-36\ne 0$$

25. Factor the numerator.

26. Expand any powers or inner brackets before factoring.

    $$9\times x+6=9\times x+6$$

27. Take the common factor out of every term. Show each remaining term after division.

    $$9\times x+6=3\times \left(3\times x+2\right)$$

28. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    $$9\times x+6=3\times \left(3\times x+2\right)$$

29. The complete factored form is:

    $$3\times \left(3\times x+2\right)$$

30. Factor the denominator.

31. Expand any powers or inner brackets before factoring.

    $$x^3+2\times x^2-15\times x-36=x^3+2\times x^2-15\times x-36$$

32. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    $$x^3+2\times x^2-15\times x-36=\left(x-4\right)\times {\left(x+3\right)}^2$$

33. The complete factored form is:

    $$\left(x-4\right)\times {\left(x+3\right)}^2$$

34. Write the factored numerator over the factored denominator.

    $$\frac{3\times \left(3\times x+2\right)}{\left(x-4\right)\times {\left(x+3\right)}^2}$$

35. Write the fraction in lowest form. Only factors, never added terms, have been cancelled.

    $$\frac{9\times x+6}{\left(x-4\right)\times {\left(x+3\right)}^2}$$

Answer

The final answer is shown below.