Q1.
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²) (ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²) (iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx) (iv) (4y² − 20yz + 25z²)/(25z² − 4y²) (v) [(x² + x − 6)(x² − 7x + 12)]/[(x² − 6x + 8)(x² − 9)] (vi) (p⁴ − 16)/(p² − 4p + 4)
Answer
The method is always the same: factor the top, factor the bottom, cancel whatever is common and non-zero.
(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)
Numerator: 3p² − 3pq − 18q² = 3(p² − pq − 6q²) = 3(p − 3q)(p + 2q)
Denominator: p² + 3pq − 10q² = (p + 5q)(p − 2q)
= 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]
Denominator: p² + 3pq − 10q² = (p + 5q)(p − 2q)
= 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]
The four factors are all different, so nothing cancels — the expression is already in its lowest terms, and the fully factorised form above is the simplest answer. (A quick check: put p = 1, q = 0. The original gives 3/1 = 3, and so does the factorised form.)
(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²) — the numerator is a cube of a difference.
Numerator = n³ − 3n²m + 3nm² − m³ = (n − m)³
Denominator = 5(m² − 2mn + n²) = 5(m − n)² = 5(n − m)²
= (n − m)³ / [5(n − m)²] = (n − m)/5
Denominator = 5(m² − 2mn + n²) = 5(m − n)² = 5(n − m)²
= (n − m)³ / [5(n − m)²] = (n − m)/5
(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx) — the numerator is the three-cubes identity with v negated.
Numerator = w³ + (−v)³ + x³ − 3(w)(−v)(x)
= (w − v + x)(w² + v² + x² + wv + vx − wx)
Denominator = (w − v + x)²
= (w² + v² + x² + wv + vx − wx)/(w − v + x)
= (w − v + x)(w² + v² + x² + wv + vx − wx)
Denominator = (w − v + x)²
= (w² + v² + x² + wv + vx − wx)/(w − v + x)
(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)
Numerator = (2y − 5z)²
Denominator = (5z)² − (2y)² = (5z + 2y)(5z − 2y)
= (2y − 5z)² / [(5z + 2y)(5z − 2y)]
and (2y − 5z) = −(5z − 2y), so one factor cancels leaving a minus sign:
= −(2y − 5z)/(2y + 5z) = (5z − 2y)/(2y + 5z)
Denominator = (5z)² − (2y)² = (5z + 2y)(5z − 2y)
= (2y − 5z)² / [(5z + 2y)(5z − 2y)]
and (2y − 5z) = −(5z − 2y), so one factor cancels leaving a minus sign:
= −(2y − 5z)/(2y + 5z) = (5z − 2y)/(2y + 5z)
(v) [(x² + x − 6)(x² − 7x + 12)] / [(x² − 6x + 8)(x² − 9)]
x² + x − 6 = (x + 3)(x − 2)
x² − 7x + 12 = (x − 3)(x − 4)
x² − 6x + 8 = (x − 2)(x − 4)
x² − 9 = (x − 3)(x + 3)
= [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)]
Every factor on top appears below → = 1
x² − 7x + 12 = (x − 3)(x − 4)
x² − 6x + 8 = (x − 2)(x − 4)
x² − 9 = (x − 3)(x + 3)
= [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)]
Every factor on top appears below → = 1
(vi) (p⁴ − 16)/(p² − 4p + 4)
Numerator = (p²)² − 4² = (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4)
Denominator = (p − 2)²
= (p + 2)(p² + 4)/(p − 2)
Denominator = (p − 2)²
= (p + 2)(p² + 4)/(p − 2)
Why cancelling needs a condition: writing (n − m)³/[5(n − m)²] = (n − m)/5 divides top and bottom by (n − m)², which is legal only when n ≠ m. The question grants exactly this by assuming the denominators are non-zero. Without that promise the two expressions would not be equal everywhere — the left side would be undefined at n = m while the right side is 0 there.
Tip: In part (iii) it pays to recognise the shape x³ + y³ + z³ − 3xyz before doing anything else. Here y = −v, which turns −3xyz into +3wvx and −v³ into the −v³ you can see. The denominator, meanwhile, is just (w − v + x)².