Mathematics 10 — Exercise 5.1 · Q1 (ix)

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.

3×x2+6×xx2+4×x+4

Step-by-step solution

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

    x2+4×x+4≠0
  2. Factor the numerator.

  3. Expand any powers or inner brackets before factoring.

    3×x2+6×x=3×x2+6×x
  4. Take the common factor out of every term. Show each remaining term after division.

    3×x2+6×x=(3×x)×(x+2)
  5. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    3×x2+6×x=3×x×(x+2)
  6. The complete factored form is:

    3×x×(x+2)
  7. Factor the denominator.

  8. Expand any powers or inner brackets before factoring.

    x2+4×x+4=x2+4×x+4
  9. Split the middle coefficient: these two numbers add to b and multiply to a·c.

    (2+2=4,2×2=4)
  10. Rewrite the middle term as those two terms.

    x2+2×x+2×x+4
  11. Group two terms at a time and take each common factor.

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

    (x+2)(x+2)
  13. The complete factored form is:

    (x+2)2
  14. Write the factored numerator over the factored denominator.

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

    x+2
  16. Divide numerator and denominator by that same nonzero factor.

    3×x2+6×xx+2x2+4×x+4x+2
  17. Simplify the two divisions separately.

    3×xx+2
  18. Write the fraction in lowest form. Only factors, never added terms, have been cancelled.

    3×xx+2

Answer

The final answer is shown below.

3×xx+2