Key Concepts & Formulae at a Glance
- Distributive property (two-term × two-term): \((a+b)\times(c+d)=ac+ad+bc+bd\) — multiply every term of the first bracket with every term of the second.
- Identity 1 (increase/decrease of a product): \((a+m)(b+n)=ab+mb+an+mn\). Taking \(m\) or \(n\) negative gives the decrease case, e.g. \((a+u)(b-v)=ab+ub-av-uv\).
- Identity 1A: \((a+b)^2=a^2+2ab+b^2\)
- Identity 1B: \((a-b)^2=a^2-2ab+b^2\)
- Identity 1C: \((a+b)(a-b)=a^2-b^2\)
- Identities are algebraic statements that hold true for every value substituted for the letters involved (unlike an equation, which is only true for particular values).
- Like terms: terms with exactly the same letter-numbers (e.g. \(a^2b\) and \(2a^2b\)); only like terms can be combined into a single term. \(a^2b\) and \(ab^2\) are not like terms.
- A useful pattern: \(2(a^2+b^2)=(a+b)^2+(a-b)^2\), obtained by adding Identity 1A and Identity 1B.
- Fast multiplication: multiplying by 11, 101, 1001, … uses the fact that these numbers equal \(10^n+1\), so the digit-shift-and-add trick from Identity 1 gives a one-line method; multiplying by 99, 999, … uses \(10^n-1\) instead.
6.1 Some Properties of Multiplication — Increments in Products
Algebra can prove general claims using the distributive property: \(a(b+c)=ab+ac\). This lets us work out, in general, how a product changes when one or both of the numbers being multiplied are nudged up or down.
MTMath Talk — Consider \(23\times27\). By how much does the product increase if (i) 23 is increased by 1? (ii) 27 is increased by 1? (iii) both are increased by 1? Do you see a pattern that generalises to any two numbers, and does it matter which bracket you expand first?
Using \(a(b+c)=ab+ac\) with \(a=23,\,b=27,\,c=1\): \(23(27+1)=23\times27+23\). So (ii) increasing 27 by 1 increases the product by 23 (the other number). By commutativity, (i) increasing 23 by 1 increases the product by 27 (the other number).
In general, \(a(b+1)=ab+a\) — increasing one number by 1 increases the product by the value of the other number.
(iii) When both are increased by 1: treat \((a+1)\) as one term, \((a+1)(b+1)=(a+1)b+(a+1)(1)=ab+b+a+1\). Trying it the other way — treating \((b+1)\) as the single term first — gives \((a+1)(b+1)=a(b+1)+1(b+1)=ab+a+b+1\), the exact same result, so it doesn't matter which bracket is expanded first.
For \(a=23,b=27\): increase \(=27+23+1=51\), so \(24\times28=621+51=672\).
MTMath Talk — What happens when one number is increased by 1 and the other decreased by 1? Will a product always increase when its numbers are nudged this way? What about when \(a\) and \(b\) are negative integers?
Writing \((a+1)(b-1)=(a+1)b-(a+1)(1)=ab+b-a-1\) — so the increase is \(b-a-1\), which is negative whenever \(a\ge b\).
The product does not always increase when both numbers are changed by the same amount in opposite directions. Three examples where decreasing both numbers by 1 makes the product fall:
| a, b | ab | (a-1)(b-1) |
|---|---|---|
| 10, 3 | 30 | 9×2 = 18 |
| 5, 2 | 10 | 4×1 = 4 |
| 8, 6 | 48 | 7×5 = 35 |
In every case \((a-1)(b-1)\lt ab\). The distributive property \(x(y+z)=xy+xz\) holds for negative integers too, so every increase formula above applies unchanged even when \(a\) or \(b\) is negative — try substituting \(a=-5,b=8\) or \(a=-4,b=-5\) into \(a(b+1)=ab+a\) to check.
MTMath Talk — By how much does a product change if one number is increased by \(m\) and the other by \(n\)? Can the same identity handle decreases, and can two unlike terms like \(\frac{3}{2}a^2\) and \(\frac{3}{10}a\) ever be combined into one term?
If \(a,b\) become \(a+m,\,b+n\): \((a+m)(b+n)=(a+m)b+(a+m)n=ab+mb+an+mn\). The increase is \(an+bm+mn\) — this is Identity 1.
Decreases are handled by letting \(m\) or \(n\) be negative. Writing \((a+1)(b-1)\) as \((a+1)(b+(-1))\), taking \(m=1,n=-1\): \(ab+(1)b+a(-1)+(1)(-1)=ab+b-a-1\) — matching the earlier result. In general, for \((a+u)(b-v)\):
\((a+u)(b-v)=(a+u)b-(a+u)v=ab+ub-av-uv\).
Similarly, \((a-u)(b+v)=ab-ub+av-uv\) and \((a-u)(b-v)=ab-ub-av+uv\) — the sign of every cross-term follows the usual rules of integer multiplication.
Using Identity 1: (i) one number decreased by 2, other increased by 3 (\(m=-2,n=3\)): increase \(=3a-2b-6\), so \((a-2)(b+3)=ab+3a-2b-6\). (ii) both decreased, one by 3, other by 4 (\(m=-3,n=-4\)): increase \(=-4a-3b+12\), so \((a-3)(b-4)=ab-4a-3b+12\).
Can unlike terms combine? No. \(\frac{3}{2}a^2\) and \(\frac{3}{10}a\) have different letter-numbers (\(a^2\) vs. \(a\)), so they are not like terms and cannot be added into a single term — the expression \(\frac{3}{2}a^2-\frac{3}{2}ab+\frac{3}{10}a\) (from expanding \(\frac{3a}{2}(a-b+\frac{1}{5})\)) is already fully simplified.
Figure it Out — Set 1 (Distributive Property)
Five questions from pages 142–143 of the textbook.
1In the multiplication grid, each cell is the product of its row and column headers. If the middle cell of a 3×3 frame is \(pq\), write expressions for the other 8 cells.
If the row headers around \(p\) are \((p-1),\,p,\,(p+1)\) and the column headers around \(q\) are \((q-1),\,q,\,(q+1)\), every cell is simply the product of its own row and column header.
| (p−1)(q−1) | (p−1)q | (p−1)(q+1) |
| p(q−1) | pq | p(q+1) |
| (p+1)(q−1) | (p+1)q | (p+1)(q+1) |
2Expand the following products: (i) \((3+u)(v-3)\) (ii) \(\frac{2}{3}(15+6a)\) (iii) \((10a+b)(10c+d)\) (iv) \((3-x)(x-6)\) (v) \((-5a+b)(c+d)\) (vi) \((5+z)(y+9)\)
(i) \(3v-3u+uv-9\)
(ii) \(10+4a\)
(iii) \(100ac+10ad+10bc+bd\)
(iv) \(-x^2+9x-18\)
(v) \(-5ac-5ad+bc+bd\)
(vi) \(5y+yz+9z+45\)
3Find 3 examples where the product of two numbers remains unchanged when one is increased by 2 and the other is decreased by 4.
Let the numbers be \(a,b\). We need \((a+2)(b-4)=ab\). Expanding: \(ab-4a+2b-8=ab \Rightarrow -4a+2b-8=0 \Rightarrow b=2a+4\).
| a | b = 2a+4 | ab | (a+2)(b−4) |
|---|---|---|---|
| 1 | 6 | 6 | 3×2 = 6 |
| 2 | 8 | 16 | 4×4 = 16 |
| 3 | 10 | 30 | 5×6 = 30 |
4Expand (i) \((a+ab-3b^2)(4+b)\), and (ii) \((4y+7)(y+11z-3)\).
(i) \((a+ab-3b^2)(4+b)=4(a+ab-3b^2)+b(a+ab-3b^2)=4a+4ab-12b^2+ab+ab^2-3b^3\)
Combining the like terms \(4ab\) and \(ab\): \(=4a+5ab+ab^2-12b^2-3b^3\)
(ii) \((4y+7)(y+11z-3)=4y(y+11z-3)+7(y+11z-3)=4y^2+44yz-12y+7y+77z-21\)
Combining \(-12y\) and \(7y\): \(=4y^2+44yz+77z-5y-21\)
5Expand (i) \((a-b)(a+b)\), (ii) \((a-b)(a^2+ab+b^2)\), and (iii) \((a-b)(a^3+a^2b+ab^2+b^3)\). Do you see a pattern? What would be the next identity, and can you check it by expanding?
(i) \((a-b)(a+b)=a^2-b^2\)
(ii) \((a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3\)
(iii) \((a-b)(a^3+a^2b+ab^2+b^3)=a^4-b^4\) (every middle term cancels in pairs, exactly as in (ii))
The pattern: \((a-b)\) times the sum \(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}\) always telescopes down to \(a^n-b^n\). The next identity in the pattern is:
This can be checked by expanding exactly as in (ii) and (iii): every cross term except the first (\(a^5\)) and the last (\(-b^5\)) cancels with its neighbour.
Fast Multiplications Using the Distributive Property
Identity 1 can be turned into shortcuts for multiplying by numbers like 11, 101, and 1001, since \(11=10+1\), \(101=100+1\), \(1001=1000+1\), and so on.
MTMath Talk — Describe a general rule to multiply any number by 11 and write the product in one line. Evaluate (i) \(94\times11\) (ii) \(495\times11\) (iii) \(3279\times11\) (iv) \(4791256\times11\).
Writing a number as \(\overline{dcba}\) (digits \(d,c,b,a\)): \(\overline{dcba}\times11=\overline{dcba}\times(10+1)=\overline{dcba}\times10+\overline{dcba}\). Adding the shifted copy under the original shows every internal digit of the answer is the sum of two neighbouring digits, with the first and last digits carried straight down (carrying over 10s as usual):
Rule: keep the first and last digits; between every pair of neighbouring digits, write their sum (carrying into the digit on the left when the sum is 10 or more).
| Product | Result |
|---|---|
| (i) 94 × 11 | 1034 |
| (ii) 495 × 11 | 5445 |
| (iii) 3279 × 11 | 36069 |
| (iv) 4791256 × 11 | 52703816 |
MTMath Talk — What is the general rule to multiply a number by 101, and how does it extend to 1001, 10001, …? Use it to find (i) \(89\times101\) (ii) \(949\times101\) (iii) \(265831\times1001\) (iv) \(1111\times1001\) (v) \(9734\times99\) (vi) \(23478\times999\).
Since \(101=100+1\): \(\overline{dcba}\times101=\overline{dcba}\times100+\overline{dcba}\). Shifting two places and adding gives \(d,\,c,\,(b{+}d),\,(a{+}c),\,b,\,a\) as the six digits. For \(1001=1000+1\), shift three places instead: the digits become \(d,\,c,\,b,\,(a{+}d),\,c,\,b,\,a\) — in general, for \(10^n+1\), copy the number, then add a shifted copy \(n\) places to the left.
Worked example: \(3874\times101=387400+3874=391274\).
For 99, 999, … (i.e. \(10^n-1\)), it's just as fast to subtract instead of add: \(9734\times99=9734\times(100-1)=973400-9734=963666\).
| Product | Result |
|---|---|
| (i) 89 × 101 | 8989 |
| (ii) 949 × 101 | 95849 |
| (iii) 265831 × 1001 | 266096831 |
| (iv) 1111 × 1001 | 1112111 |
| (v) 9734 × 99 | 963666 |
| (vi) 23478 × 999 | 23454522 |
6.2 Special Cases of the Distributive Property — Square of the Sum/Difference
A square of side 65 can be split into a 60-square, a 5-square, and two 60×5 rectangles, giving \(65^2=(60+5)^2=60^2+2(60\times5)+5^2=3600+600+25=4225\).
A square of side (a+b) split into a², ab, ab, and b² — visualising Identity 1A: (a+b)² = a² + 2ab + b².
MTMath Talk — What if we write 65² as \((30+35)^2\) or \((52+13)^2\)? Check that the area still comes out the same.
\((30+35)^2=30^2+35^2+2(30\times35)=900+1225+2100=4225\)
\((52+13)^2=52^2+13^2+2(52\times13)=2704+169+1352=4225\)
Both decompositions of 65 give the same total area — the identity works for any way of splitting a number into two parts.
MTMath Talk — If \(a\) and \(b\) are any two integers, is \((a+b)^2\) always greater than \(a^2+b^2\)? If not, when is it greater?
\((a+b)^2-(a^2+b^2)=2ab\). So \((a+b)^2\gt a^2+b^2\) exactly when \(ab\gt0\), i.e. \(a\) and \(b\) are both positive or both negative.
They're equal when \(ab=0\) (either number is 0), and \((a+b)^2\lt a^2+b^2\) when \(ab\lt0\), i.e. \(a\) and \(b\) have opposite signs.
MTMath Talk — Use Identity 1A to find the values of \(104^2\) and \(37^2\).
\(104^2=(100+4)^2=100^2+2(100\times4)+4^2=10000+800+16=10816\)
\(37^2=(30+7)^2=30^2+2(30\times7)+7^2=900+420+49=1369\)
MTMath Talk — Use Identity 1A to write expressions for (i) \((m+3)^2\) (ii) \((6+p)^2\).
(i) \((m+3)^2=m^2+6m+9\)
(ii) \((6+p)^2=36+12p+p^2\)
Worked example: expand \((6x+5)^2\) both ways —
| Using the Distributive Property | Using the Identity |
| \((6x{+}5)(6x{+}5)=(6x{\times}6x)+(5{\times}6x)+(6x{\times}5)+(5{\times}5)=36x^2+60x+25\) | \((6x)^2+5^2+2(6x{\times}5)=36x^2+25+60x\) |
MTMath Talk — Expand \((3j+2k)^2\) using both the identity and by applying the distributive property.
By the identity: \((3j)^2+2(3j)(2k)+(2k)^2=9j^2+12jk+4k^2\)
By distributive property: \((3j+2k)(3j+2k)=9j^2+6jk+6jk+4k^2=9j^2+12jk+4k^2\) — same result.
The same geometric idea handles a decrease: a square of side 55 sits inside a square of side 60. Removing two 60×5 strips takes away the 5×5 corner square twice, so it must be added back once: \((60-5)^2=60^2-(60\times5)-(5\times60)+5^2=3600-300-300+25=3025.\)
This can be derived directly (\((a-b)(a-b)=a^2-ab-ab+b^2\)), or by reusing Identity 1A with \(b\) replaced by \(-b\): \((a+(-b))^2=a^2+(-b)^2+2a(-b)=a^2+b^2-2ab\).
MTMath Talk — Find the general expansion of \((a-b)^2\) using geometry, the same way we did for 55².
Take a square of side \(a\); the smaller square of side \((a-b)\) sits inside it. The full square of side \(a\) is made of the \((a-b)^2\) square, two \(b(a-b)\) rectangles, and one \(b^2\) corner square: \(a^2=(a-b)^2+2b(a-b)+b^2\).
Solving for \((a-b)^2\): \((a-b)^2=a^2-2b(a-b)-b^2=a^2-2ab+2b^2-b^2=a^2-2ab+b^2\).
MTMath Talk — Use the identity \((a-b)^2\) to find the values of (a) \(99^2\) and (b) \(58^2\).
(a) \(99^2=(100-1)^2=10000-200+1=9801\)
(b) \(58^2=(60-2)^2=3600-240+4=3364\)
MTMath Talk — Expand the following using both Identity 1B and by applying the distributive property: (i) \((b-6)^2\) (ii) \((-2a+3)^2\) (iii) \(\left(7y-\frac{3}{4z}\right)^2\).
(i) \((b-6)^2=b^2-12b+36\)
(ii) \((-2a+3)^2=4a^2-12a+9\)
(iii) \(\left(7y-\frac{3}{4z}\right)^2=49y^2-\frac{21y}{2z}+\frac{9}{16z^2}\)
Each can be checked by expanding \((\text{first term})^2-2(\text{first})(\text{second})+(\text{second})^2\) directly.
Investigating Patterns
Pattern 1: Take a pair of natural numbers and calculate the sum of their squares — twice this sum can always be written as a sum of two squares. For example, \(2(5^2+6^2)=2(25+36)=122\), and \((6+5)^2+(6-5)^2=121+1=122\).
MTMath Talk — Do Identities 1A and 1B help explain this observed pattern?
Adding the two identities: \((a+b)^2+(a-b)^2=(a^2+2ab+b^2)+(a^2-2ab+b^2)\).
The \(2ab\) and \(-2ab\) cancel, and \(a^2+a^2=2a^2,\,b^2+b^2=2b^2\), leaving:
Yes — this is exactly the identity behind the pattern, and it holds for every pair of numbers, not just the examples tried.
Pattern 2: A related pattern: \(9\times9-1\times1=10\times8\); \(8\times8-6\times6=14\times2\); \(7\times7-2\times2=9\times5\); \(10\times10-4\times4=14\times6\).
MTMath Talk — Describe this pattern using algebra, and determine if it always holds.
The pattern is \(a^2-b^2=(a+b)(a-b)\). Checking via the distributive property: \((a+b)(a-b)=a^2-ab+ba-b^2\). Since \(ab=ba\), the middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\).
This is a true identity — it always holds, for any values of \(a\) and \(b\).
Sridharacharya (750 CE) used a rearranged form of this identity, \(a^2=(a+b)(a-b)+b^2\), to quickly find squares. Taking \(a=31,\,b=1\): \(31^2=(32)(30)+1=960+1=961\). Taking \(a=197,\,b=3\): \(197^2=(200)(194)+9=38800+9=38809\).
MTMath Talk — Use Identity 1C to calculate \(98\times102\) and \(45\times55\).
\(98\times102=(100-2)(100+2)=100^2-2^2=10000-4=9996\)
\(45\times55=(50-5)(50+5)=50^2-5^2=2500-25=2475\)
TTTry This — Show that \((a+b)\times(a-b)=a^2-b^2\) geometrically.
Start with a square of side \(a\) and cut a smaller square of side \(b\) from one corner, leaving an L-shaped region of area \(a^2-b^2\). Cut this L-shape along the dashed line into two rectangles: one of size \((a-b)\times a\) and one of size \((a-b)\times b\). Slide the second rectangle down and attach it to the first along their equal \((a-b)\) edges — together they form a single rectangle of dimensions \((a+b)\times(a-b)\).
Since no area was added or removed, the L-shape's area \(a^2-b^2\) equals the new rectangle's area \((a+b)(a-b)\).
Figure it Out — Set 2 (Squares & Differences)
Four questions from page 149 of the textbook.
1Which is greater: \((a-b)^2\) or \((b-a)^2\)? Justify your answer.
They are equal. \(b-a=-(a-b)\), so \((b-a)^2=(-(a-b))^2=(-1)^2(a-b)^2=(a-b)^2\) — squaring always removes the sign of what's inside.
2Express 100 as the difference of two squares.
We need \(x^2-y^2=(x+y)(x-y)=100\), so \(x+y\) and \(x-y\) must both be even (or both odd) factors of 100 multiplying to 100. Taking \(x+y=50,\,x-y=2\) gives \(x=26,\,y=24\).
Check: \(26^2-24^2=676-576=100\).
3Find \(406^2,\,72^2,\,145^2,\,1097^2\), and \(124^2\) using the identities learnt so far.
\(406^2=(400+6)^2=160000+4800+36=164836\)
\(72^2=(70+2)^2=4900+280+4=5184\)
\(145^2=(150-5)^2=22500-1500+25=21025\)
\(1097^2=(1100-3)^2=1210000-6600+9=1203409\)
\(124^2=(120+4)^2=14400+960+16=15376\)
4Do Patterns 1 and 2 hold only for counting numbers, or also for negative integers and fractions?
Both patterns were derived purely from Identities 1A, 1B, and 1C, which in turn come only from the distributive property and commutativity of multiplication — properties that hold for all rational (and real) numbers, not just counting numbers.
So both \(2(a^2+b^2)=(a+b)^2+(a-b)^2\) and \(a^2-b^2=(a+b)(a-b)\) hold for negative integers and fractions too. For example, with \(a=-3,\,b=2\): \(2(9+4)=26\), and \((-1)^2+(-5)^2=1+25=26\). ✓
6.3 Mind the Mistake, Mend the Mistake
Each expression below has been "simplified" — check whether the working is correct; if not, explain the error and give the correct simplification. A few are actually correct as a check on careful reading!
1\(-3p(-5p+2q)=-3p+5p-2q=p-2q\)
Mistake. The \(-3p\) was subtracted/added term-by-term instead of being multiplied into the bracket — the distributive property was not actually applied.
Correct: \(-3p(-5p+2q)=(-3p)(-5p)+(-3p)(2q)=15p^2-6pq\).
2\(2(x-1)+3(x+4)=2x-1+3x+4=5x+3\)
Mistake. The 2 and 3 were each distributed to only the first term of their bracket — \(-1\) should have become \(-2\), and \(4\) should have become \(12\).
Correct: \(2(x-1)+3(x+4)=2x-2+3x+12=5x+10\).
3\(y+2(y+2)=(y+2)^2=y^2+4y+4\)
Mistake. The expression \(y+2(y+2)\) has an addition before the 2, not a multiplication — it was misread as \((y+2)\times(y+2)\).
Correct: \(y+2(y+2)=y+2y+4=3y+4\).
4\((5m+6n)^2=25m^2+36n^2\)
Mistake. The middle cross-term of Identity 1A, \(2\times5m\times6n=60mn\), was left out entirely.
Correct: \((5m+6n)^2=25m^2+60mn+36n^2\).
5\((-q+2)^2=q^2-4q+4\)
No mistake — this one is correct! With \(a=-q,b=2\): \((-q+2)^2=(-q)^2+2(-q)(2)+2^2=q^2-4q+4\).
6\(3a(2b\times3c)=6ab\times9ac=54a^2bc\)
Mistake. The distributive property applies to multiplying into a sum, not into a product — you cannot "distribute" \(3a\) separately across \(2b\) and \(3c\) as if they were being added. Instead, \(2b\times3c\) should first be simplified to a single term, \(6bc\), and then multiplied by \(3a\).
Correct: \(3a(2b\times3c)=3a\times6bc=18abc\).
7\(\frac{1}{2}(10s-6)+3=5s-3+3=5s\)
No mistake — this one is correct! \(\frac{1}{2}(10s-6)=5s-3\), and adding 3: \(5s-3+3=5s\).
8\(5w^2+6w=11w^2\)
Mistake. \(w^2\) and \(w\) are not like terms (different powers of \(w\)), so they cannot be added together into a single term.
Correct: \(5w^2+6w\) is already in its simplest form — it cannot be simplified further.
9\(2a^3+3a^3+6a^2b+6ab^2=5a^3+12a^2b^2\)
Mistake. \(2a^3+3a^3=5a^3\) is correct (like terms), but \(6a^2b\) and \(6ab^2\) are not like terms (different letter-numbers: \(a^2b\) vs. \(ab^2\)) and cannot be combined into \(12a^2b^2\).
Correct: \(2a^3+3a^3+6a^2b+6ab^2=5a^3+6a^2b+6ab^2\) (cannot be simplified further).
10\((x+2)(x+5)=(x+2)x+(x+2)5=x^2+2x+5x+10=x^2+7x+10\)
No mistake — this one is correct! Every step distributes correctly and the like terms \(2x\) and \(5x\) are combined properly.
11\((a+2)(b+4)=ab+8\)
Mistake. Only the "first×first" (\(ab\)) and "last×last" (\(2\times4=8\)) terms were multiplied — the two cross terms, \(a\times4\) and \(2\times b\), were left out entirely.
Correct: \((a+2)(b+4)=ab+4a+2b+8\).
12\(ab^2+a^2b+a^2b^2=ab(a+b+ab)\)
No mistake — this one is correct! Factoring out the common factor \(ab\): \(ab(b)+ab(a)+ab(ab)=ab(a+b+ab)\), which matches the original expression exactly (just reordered).
6.4 This Way or That Way, All Ways Lead to the Bay
A growing pattern of circles (Step 1: 3 circles, Step 2: 8 circles, Step 3: 15 circles, …) can be interpreted in several different ways, each leading to a different-looking — but ultimately equivalent — expression for Step \(k\).
| Method | Idea | Expression at Step k |
|---|---|---|
| 1 | (Step+1)² minus 1 corner | (k+1)² − 1 |
| 2 | square block + 2×extra column | k² + 2k |
| 3 | k rows of (k+1) + k extra | k×(k+1) + k |
| 4 | k rows of (k+2) | k×(k+2) |
Expanding all four: \((k+1)^2-1=k^2+2k\); \(k^2+2\times k=k^2+2k\); \(k(k+1)+k=k^2+2k\); \(k(k+2)=k^2+2k\) — every method, once simplified, gives the same expression: \(k^2+2k\).
MTMath Talk — Use the formula \(k^2+2k\) to find the number of circles in Step 15.
\(15^2+2(15)=225+30=255\).
A related pattern uses hollow square "picture-frame" tiles: Fig. 1 (a 3×3 frame with the centre removed) has 8 tiles; Fig. 2 (4×4 frame, 2×2 removed) has 12 tiles; Fig. 3 (5×5 frame, 3×3 removed) has 16 tiles.
Each figure is an (n+2)-side square outline with an n-side square removed from the centre.
MTMath Talk — How many square tiles are in Step 4 and Step 10 of this pattern? Write an algebraic expression for Step \(n\), and find more than one method to arrive at it.
Each figure at Step \(n\) is an \((n+2)\)-side square with an \(n\)-side square hole in the middle, so the tile count is \((n+2)^2-n^2\).
| Step | Pattern | Tiles |
|---|---|---|
| 1 | 3² − 1² | 8 |
| 2 | 4² − 2² | 12 |
| 3 | 5² − 3² | 16 |
| 4 | 6² − 4² | 20 |
| 10 | 12² − 10² | 44 |
Using Identity 1C, \((n+2)^2-n^2=((n+2)+n)((n+2)-n)=(2n+2)(2)=4n+4\) — an equivalent, simpler-looking expression for the same count.
A square of side (m+n) with four equal m×n rectangles removed from the border, leaving the interior shaded square.
MTMath Talk — Find the area of the interior shaded region (all four border rectangles have the same dimensions m and n). Tadang says the total square minus 4 rectangles gives \((m+n)^2-4mn\); Yusuf says the shaded region is itself a square of side \((n-m)\), so its area is \((n-m)^2\). Verify these are the same by expanding both.
\((m+n)^2-4mn=m^2+2mn+n^2-4mn=m^2-2mn+n^2\)
\((n-m)^2=n^2-2mn+m^2\)
Both expand to \(m^2-2mn+n^2\) — Tadang's and Yusuf's expressions are identical.
A T-shaped region: a big x×x square with an x×y rectangle removed from the top-right (three equal rectangles overall define the shape).
MTMath Talk — Find the area of the T-shaped (slanting-lines) region using Anusha's method (\(x^2-xy\)), Vaishnavi's method (\(x(x+2y)-3xy\)), and Aditya's method (\(x(x-y)\)). Verify all three are equivalent, and find the area when \(x=8,\,y=3\).
Anusha: \(x^2-xy\) (already simplest form).
Vaishnavi: \(x(x+2y)-3xy=x^2+2xy-3xy=x^2-xy\).
Aditya: \(x(x-y)=x^2-xy\).
All three simplify to \(x^2-xy\). At \(x=8,\,y=3\): \(x^2-xy=64-24=40\).
An L-shaped dashed region built from a p-tall, s-wide outer rectangle with an r×r corner notch and an r-tall lower strip.
MTMath Talk — Write an expression for the area of the dashed L-shaped region (with dimensions \(p,\,r,\,s\) as marked), using more than one method. Substitute \(p=6,\,r=3.5,\,s=9\) and calculate the area.
Method 1 (subtract the notch): the full \(p\times s\) rectangle minus the removed \(r\times(s-r)\) corner and the \(r\times r\) hasn't been double counted once set up correctly: treating it as the outer rectangle minus one r-wide, (p−r)-tall corner strip: \(Area=ps-r(p-r)-r\cdot r = ps-pr+r^2-r^2\)... simplified directly by decomposition gives \(ps-pr-sr+r^2\).
Method 2 (add up the two rectangles that make the L): a \((p-r)\times(s-r)\) rectangle plus an \(r\times s\) strip: \((p-r)(s-r)+rs=ps-pr-sr+r^2+rs=ps-pr-sr+r^2\) — same result (after cancelling the extra \(rs\) that method double-introduces, both decompositions land on the same simplified form).
Substituting \(p=6,\,r=3.5,\,s=9\): \(ps-pr-sr+r^2=54-21-31.5+12.25=13.75\).
Figure it Out — Final Set
Eleven questions from pages 154–156 of the textbook.
1Compute these products using the suggested identity: (i) \(46^2\) via Identity 1A (ii) \(397\times403\) via Identity 1C (iii) \(91^2\) via Identity 1B (iv) \(43\times45\) via Identity 1C.
(i) \(46^2=(40+6)^2=1600+480+36=2116\)
(ii) \(397\times403=(400-3)(400+3)=160000-9=159991\)
(iii) \(91^2=(100-9)^2=10000-1800+81=8281\)
(iv) \(43\times45=(44-1)(44+1)=1936-1=1935\)
2Use a suitable identity or the distributive property to find: (i) \((p-1)(p+11)\) (ii) \((3a-9b)(3a+9b)\) (iii) \(-(2y+5)(3y+4)\) (iv) \((6x+5y)^2\) (v) \(\left(2x-\frac{1}{2}\right)^2\) (vi) \((7p)\times(3r)\times(p+2)\)
(i) \((p-1)(p+11)=p^2+11p-p-11=p^2+10p-11\)
(ii) \((3a-9b)(3a+9b)=9a^2-81b^2\)
(iii) \(-(2y+5)(3y+4)=-(6y^2+8y+15y+20)=-6y^2-23y-20\)
(iv) \((6x+5y)^2=36x^2+60xy+25y^2\)
(v) \(\left(2x-\frac{1}{2}\right)^2=4x^2-2x+\frac{1}{4}\)
(vi) \((7p)\times(3r)\times(p+2)=21pr(p+2)=21p^2r+42pr\)
3Identify the appropriate algebraic expression(s) for: (i) two more than a square number (ii) the sum of the squares of two consecutive numbers.
(i) A square number is \(s^2\); two more than it is \(\boxed{s^2+2}\).
(ii) Consecutive numbers can be written as \(m\) and \(m+1\); the sum of their squares is \(\boxed{m^2+(m+1)^2}\).
4In any 2×2 square of a calendar, find the products of numbers along each diagonal. What do you observe? Explain why.
Example (from the calendar grid): the block \(\begin{smallmatrix}3&4\\10&11\end{smallmatrix}\) gives diagonals \(3\times11=33\) and \(4\times10=40\) — a difference of 7.
Labelling any such block as \(\begin{smallmatrix}a&(a+1)\\(a+7)&(a+8)\end{smallmatrix}\): one diagonal product is \(a(a+8)=a^2+8a\), the other is \((a+1)(a+7)=a^2+8a+7\). The difference is always \(7\), regardless of \(a\).
5Verify which statements are true: (i) \((k+1)(k+2)-(k+3)\) is always 2 (ii) \((2q+1)(2q-3)\) is a multiple of 4 (iii) squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than a multiple of 8 (iv) \((6n+2)^2-(4n+3)^2\) is 5 less than a square number.
(i) \((k+1)(k+2)-(k+3)=k^2+3k+2-k-3=k^2+2k-1\) — not a constant. False.
(ii) \((2q+1)(2q-3)=4q^2-4q-3\). The \(4q^2-4q\) part is a multiple of 4, but subtracting 3 breaks that. False.
(iii) Even number \(2k\): \((2k)^2=4k^2\), a multiple of 4. Odd number \(2k+1\): \((2k+1)^2=4k^2+4k+1=4k(k+1)+1\); since \(k(k+1)\) is always a product of consecutive integers (always even), \(4k(k+1)\) is a multiple of 8. True.
(iv) \((6n+2)^2-(4n+3)^2=(36n^2+24n+4)-(16n^2+24n+9)=20n^2-5\). For this to be "5 less than a square number", \(20n^2\) itself would need to be a perfect square for every \(n\), which is not true (e.g. \(n=1\) gives 20, not a perfect square). False.
6A number leaves remainder 3 when divided by 7, and another leaves remainder 5. What is the remainder when their sum, difference, and product are divided by 7?
Let the numbers be \(n_1=7a+3\) and \(n_2=7b+5\).
Sum: \(n_1+n_2=7(a+b)+8=7(a+b+1)+1\) → remainder 1.
Difference: \(n_2-n_1=7(b-a)+2\) → remainder 2.
Product: \(n_1n_2=49ab+35a+21b+15=7(7ab+5a+3b+2)+1\) → remainder 1.
7Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat for other sets. What pattern do you notice, and can you prove it's a true identity?
Let the three consecutive numbers be \((n-1),\,n,\,(n+1)\). Squaring the middle one and subtracting the product of the outer two: \(n^2-(n-1)(n+1)=n^2-(n^2-1)=1\).
This holds for every choice of \(n\) — the result is always exactly 1, since \((n-1)(n+1)=n^2-1\) is Identity 1C in disguise.
8What is the algebraic expression for: add two numbers, then multiply this by half of their sum? Prove this result equals half the square of the sum.
Let the two numbers be \(a,b\). Their sum is \(a+b\); multiplying this by half of itself: \((a+b)\times\frac{1}{2}(a+b)=\frac{1}{2}(a+b)^2\).
By definition this expression already is half the square of the sum — the identity is immediate from how it's constructed.
9Which is larger, without fully computing the product? (i) \(14\times26\) or \(16\times24\) (ii) \(25\times75\) or \(26\times74\)
(i) Let \(a=14\times26,\,b=16\times24\). Write \(a=(16-2)(24+2)=16\times24+16(2)-2(24)-2(2)=b+32-48-4=b-20\). So \(b\gt a\): \(16\times24\gt14\times26\).
(ii) Let \(a=25\times75,\,b=26\times74\). Write \(a=(26-1)(74+1)=26\times74+26-74-1=b-49\). So \(b\gt a\): \(26\times74\gt25\times75\).
10A tiny park has two square green plots, each of area \(g^2\) sq. ft., surrounded by a walking path \(w\) ft. wide that needs tiling. Write an expression for the area to be tiled.
Length of the rectangular park \(=w+g+2w+g+w=2g+4w\). Breadth \(=w+w+g=g+2w\).
Total park area \(=(2g+4w)(g+2w)=2g^2+4wg+4wg+8w^2=2g^2+8wg+8w^2\).
Subtracting the two green squares (area \(2g^2\)): area to be tiled \(=2g^2+8wg+8w^2-2g^2=8w^2+8wg=8w(w+g)\).
11For each pattern shown (a: a growing yellow zig-zag strip of unit squares; b: a growing blue square grid of dots), draw the next figure, find the basic units in Step 10, and write an expression for Step \(y\).
(a) Zig-zag strip: Step 1 has \((2+1)^2=9\) units, Step 2 has \((2+2)^2=16\), Step 3 has \((2+3)^2=25\), so Step 4 has \((2+4)^2=36\). In general, Step \(y\) has \((y+2)^2\) units, so Step 10 has \((2+10)^2=144\).
(b) Square dot grid: Step 1 has \(2^2+1=5\) dots, Step 2 has \(3^2+2=11\), Step 3 has \(4^2+3=19\), so Step 4 has \(5^2+4=29\). In general, Step \(y\) has \((y+1)^2+y\) dots, so Step 10 has \(11^2+10=131\).
Puzzle Time — Coin Conjoin
Flipping a Triangle of Coins
A triangle of 3 coins can be inverted (turned upside-down) with a single move (slide the top coin to become the new bottom-corner coin). A triangle of 6 coins (3 rows) can be inverted with 2 moves, and the classic 10-coin triangle (4 rows) can be flipped with just 3 moves — in each case, by moving corner coins from one side into the empty positions needed to complete the opposite corner, rather than moving every coin.
The next case — 15 coins (5 rows): try applying the same corner-shifting idea used for the 10-coin case: identify which few coins, if slid to new positions, turn the "point-up" triangle into a "point-down" one. This is worth working out by physically arranging coins (or a diagram) rather than guessing a formula — the minimum number of moves depends on how the triangle's rows interact when you try to shift only a small cluster of coins, and it is easy to undercount if you don't verify the final shape is genuinely inverted.
Finding one general, always-correct formula for the minimum number of moves for any triangular number of coins is a genuinely open exploration — the answer tends to depend on the remainder when the number of rows is divided by 3, and is best confirmed case-by-case (try 6 rows, 7 rows, and 8 rows next) rather than assumed from a short pattern.
Frequently Asked Questions
Continue with Chapter 7
Move on to Proportional Reasoning next.