Exploring Algebraic Identities

Use (A + B)² = A² + 2AB + B², taking the entire term as A or B.

(i) A = 7x, B = 4y; (ii) A = 7x/5, B = 3y/2; (iii) A = 2.5p, B = 1.5q.

(iv) A = 3s/4, B = 8t. In (v), 2x/(2y) = x/y, not x/(2y). Denominators must be nonzero.

Use (60 + 4)², (100 + 5)² and (200 + 5)².

Add base², twice base×offset, and offset².

Match the first and last terms to A² and B², then check the middle is 2AB.

(v) Take out 1/3 to leave 9a² + 12ab + 4b².

(vi) Take out 1/5 to leave 9s² + 30sv + 25v².

Use (80 − 1)², (200 − 7)² and (300 − 1)² with A² − 2AB + B².

Use 117 = 100 + 17, 78 = 80 − 2, 198 = 200 − 2.

Use 214 = 200 + 14, 1104 = 1100 + 4 and 1120 = 1100 + 20.

Expand each binomial square, including the twice-product term.

Identify the signed components of each square. In (iii), they are m/3, k/2 and 3n; their pairwise twice-products reproduce all cross terms.

In (iv), use A = p/4 and B = 4/p; AB = 1. Require p ≠ 0.

Use A² + B² + C² + 2AB + 2BC + 2CA.

Retain the negative sign in B = −2y for (ii).

Set a = 1, b = 0, c = 0: the left side is 3 and the right side is 2.

A single counterexample disproves an identity.

Choose factor pairs whose product gives the constant and whose cross terms give the linear coefficient.

For (iii), (2x − 3)(5x + 2) expands to 10x² − 11x − 6.

Use nearby-base squares for (i), (ii), (iv), (v), (viii).

For (iii), (20 + 3)(20 − 3) = 400 − 9. For (vi), (20 − 2)(30 − 1) = 600 − 20 − 60 + 2.

For (vii), (40 − 6)(40 + 3) = 1600 − 120 − 18.

Recognise signed square terms and their cross terms for (i), (ii), (iv), (v).

For (iii), use numbers −7 and +6 whose sum is −1 and product is −42.

Numerator = 3(p − 3q)(p + 2q); denominator = (p + 5q)(p − 2q).

There is no common linear factor to cancel. Retain the original nonzero-denominator restriction.

Numerator = (n − m)³; denominator = 5(n − m)².

Use A = w, B = −v, C = x. The numerator equals (w − v + x)(w² + v² + x² + wv + vx − wx).

The printed denominator is (w − v + x)²; it is not the quadratic factor in that cube identity.

Numerator = (2y − 5z)²; denominator = (5z − 2y)(5z + 2y).

Since 2y − 5z = −(5z − 2y), cancellation leaves a minus sign.

Numerator factors: (x + 3)(x − 2)(x − 3)(x − 4).

Denominator factors: (x − 2)(x − 4)(x − 3)(x + 3).

p⁴ − 16 = (p² + 4)(p − 2)(p + 2).

The denominator is (p − 2)². Cancel just one factor p − 2.

Use (A+B)²=A²+2AB+B², (A+B)(A−B)=A²−B² and (A−B)³=A³−3A²B+3AB²−B³.

In (iii), the first component is p², so its square is p⁴. In (vi), use A=1/(2r), B=4r; the twice-product is 4.

For (vii), expand the three signed terms −3m,4k,−l; for (ix), use A=7k/2 and B=2m/3.

For the first three, use difference-of-squares forms 19² − 2², 100² − 4², 20² − 4².

For cubes, use (a ± b)³ = a³ ± 3a²b + 3ab² ± b³ with nearby round bases.

(i) Match (2y + 1/(4y))²; (ii) use a difference of squares; (iii) use a difference of cubes.

For (iv), split 5x/6 as x/2 + x/3. For (v), recognise the cube of 3u − 1/5.

(vi) Use the sum of cubes of 4y and z/5. (vii) Use A = p, B = 3q, C = r in A³ + B³ + C³ − 3ABC.

(viii) is the square of 3m − 2. (ix) is one third of the same three-cubes identity with A = 3x, B = −2y, C = z.

The PDF prints 24xy and 12xz. The square (2x + 3y + 6z)² would instead have 12xy and 24xz, so it is not equal to the printed expression.

Completing the square gives (2x + 6y + 3z)² − 27(y² − z²).

If y² ≥ z², factor this as a difference of squares with A = 2x + 6y + 3z and B = 3√(3(y² − z²)).

Use (A − B)³ with A = 3u and B = 1/6.

3A²B = 9u²/2 and 3AB² = u/4.

(i) Numerator = (2x + 1)²; denominator = (2x − 1)(2x + 1).

(ii) Numerator = 27(a − 2b)(a² + 2ab + 4b²), denominator = 9(a − 2b)(a + 2b).

(iii) Numerator = (s + 5t)(s² − 5st + 25t²), denominator = (s − 7t)(s + 5t).

Factor each area expression. Use positive factors as side lengths in the geometric setting.

(i) 6a² − 24b² = 6(a − 2b)(a + 2b).

(ii) 3ps² − 15ps + 12p = 3p(s − 1)(s − 4).

Taking the path outside the playground, the outer side is 40 + 2s.

Path area = (40 + 2s)² − 40² = 160s + 4s².

If the intended path is inside instead, the area is 40² − (40 − 2s)² = 160s − 4s², with 0 ≤ s ≤ 20.

x + 1/x = 10/3, with x ≠ 0. Multiply by 3x: 3x² − 10x + 3 = 0.

Factor as (3x − 1)(x − 3) = 0.

2x² + 7x + 3 = (2x + 1)(x + 3). Divide by the nonzero width.

A quadratic with these roots has form p(x − 2)(x − 1/2) = p[x² − (5/2)x + 1].

Comparing the linear coefficient: −(5/2)p = 5, so p = −2. Comparing constants gives r = p.

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) = 25 − 20 = 5.

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 5(5 − 10).

n³ − n = n(n − 1)(n + 1), a product of three consecutive integers.

At least one is even, and one is divisible by 3. Their product is therefore divisible by 6.

Use A³+B³+C³−3ABC = (A+B+C)(A²+B²+C²−AB−BC−CA).

(i) Set A=x, B=y, C=4. The condition x+y=−4 makes the first factor zero.

(ii) Set A=x, B=−2y, C=−6. The condition x=2y+6 again makes the first factor zero.

For f(n) = n², the second difference is 2 and the third difference is 0.

Expand (n + 3)² − 3(n + 2)² + 3(n + 1)² − n² to verify all terms cancel.

The difference is (a + b)² − (a² + b²) = 2ab.

A negative difference means opposite nonzero signs; a positive difference means the same nonzero sign.

Substitute −b for b in the sum-square identity.

The three diagonal regions have areas a², b², c².

The off-diagonal regions come in equal pairs: ab, ab; ac, ac; bc, bc.

Write the number as 10k + 5. Its square is 100k(k + 1) + 25.

The upper row includes the four sign-pattern squares; the lower row has squares of sides 2a, 2b, 2c.

Expanding all four squares cancels the cross terms.

A split 2x + 5x creates a 2-by-5 corner requiring 10 unit tiles, not 12.

For an integer-sided arrangement of these tiles, the split numbers must add to 7 and multiply to 12.

An (x + 2)-by-(x + 3) rectangle contains x², five x-tiles and six units.

For x² + 11x + 30, arrange five and six x-tiles along adjacent sides and a 5-by-6 block of units.

Distribute every term from the first factor over every term of the second.

Both regroup the same product, using distributivity or associativity.

Either method yields a³ − a²b − ab² + b³.

(x − y)(x² + xy + y²) = x³ − y³; (x + y)(x² − xy + y²) = x³ + y³.

x⁴ − y⁴ = (x − y)(x + y)(x² + y²).

x⁵ − y⁵ = (x − y)(x⁴ + x³y + x²y² + xy³ + y⁴).

Numerator = (6s − t)² = (t − 6s)²; denominator = (t + 8s)(t − 6s).

Use an x-by-x corner, 3 strips along one side and 5 along the other, with a 3-by-5 unit block completing the corner.