Mathematics 10 — Exercise 5.1 · Q1 (ii)

Full question

Reduce the following algebraic fraction to its lowest terms. State the values of the variables for which the original denominator is zero and exclude them.

a×b×x+b×x2a×c×x+c×x2

Step-by-step solution

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

    a×c×x+c×x2≠0
  2. Factor the numerator.

  3. Expand any powers or inner brackets before factoring.

    a×b×x+b×x2=a×b×x+b×x2
  4. Take the common factor out of every term. Show each remaining term after division.

    a×b×x+b×x2=(b×x)×(x+a)
  5. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    a×b×x+b×x2=b×x×(a+x)
  6. The complete factored form is:

    b×x×(a+x)
  7. Factor the denominator.

  8. Expand any powers or inner brackets before factoring.

    a×c×x+c×x2=a×c×x+c×x2
  9. Take the common factor out of every term. Show each remaining term after division.

    a×c×x+c×x2=(c×x)×(x+a)
  10. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    a×c×x+c×x2=c×x×(a+x)
  11. The complete factored form is:

    c×x×(a+x)
  12. Write the factored numerator over the factored denominator.

    b×x×(a+x)c×x×(a+x)
  13. This is the common factor of numerator and denominator.

    a×x+x2
  14. Divide numerator and denominator by that same nonzero factor.

    a×b×x+b×x2a×x+x2a×c×x+c×x2a×x+x2
  15. Simplify the two divisions separately.

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

    bc

Answer

The final answer is shown below.

bc