Q1.
Try to simplify the following rational expression: (36s² − 12st + t²)/(t² + 2ts − 48s²) = (6s − t)²/[(___ + ___)(___ + ___)]. (Hint: Factor t² + 2ts − 48s² and simplify the rational expressions assuming that t² + 2ts − 48s² ≠ 0.)
Answer
Numerator. It is already a perfect square, which is why the book has written it for you:
36s² − 12st + t² = (6s)² − 2(6s)(t) + t² = (6s − t)²
Denominator. Treat it as a quadratic in t: find two terms whose sum is 2s and whose product is −48s². They are 8s and −6s.
t² + 2ts − 48s² = t² + 8ts − 6ts − 48s²
= t(t + 8s) − 6s(t + 8s)
= (t + 8s)(t − 6s)
= t(t + 8s) − 6s(t + 8s)
= (t + 8s)(t − 6s)
So the blanks are filled by (t + 8s)(t − 6s), and now the fraction cancels:
(36s² − 12st + t²)/(t² + 2ts − 48s²) = (6s − t)²/[(t + 8s)(t − 6s)]
Note (6s − t)² = (t − 6s)², since squaring kills the sign
= (t − 6s)²/[(t + 8s)(t − 6s)]
= (t − 6s)/(t + 8s)
Note (6s − t)² = (t − 6s)², since squaring kills the sign
= (t − 6s)²/[(t + 8s)(t − 6s)]
= (t − 6s)/(t + 8s)
Why the sign flip is safe: (6s − t) and (t − 6s) are negatives of each other, and a square does not notice the difference: (−u)² = u². Rewriting the numerator as (t − 6s)² makes the common factor with the denominator visible. Cancelling it is allowed because we are told t² + 2ts − 48s² ≠ 0, so in particular t − 6s ≠ 0.
Tip: When you cancel in a rational expression, always say why the factor is not zero. That is what the condition in the question is for.