We Distribute, Yet Things Multiply Class 8 Solutions

The original product is (621). Increasing 23 gives (24\times27=648), an increase of 27. Increasing 27 gives (23\times28=644), an increase of 23. Increasing both gives (24\times28=672), an increase of 51. In general, (a(b+1)-ab=a), ((a+1)b-ab=b), and ((a+1)(b+1)-ab=a+b+1).

Distribute in the other order: ((a+1)(b+1)=a(b+1)+1(b+1)=ab+a+b+1). This is the same result as distributing ((a+1)) first.

The new product is ((a+1)(b-1)=ab+b-a-1), so the change is (b-a-1). It decreases whenever (b<a+1). Examples: (10\times3=30) becomes (11\times2=22); (8\times5=40) becomes (9\times4=36); (6\times2=12) becomes (7\times1=7).

Yes. Integers satisfy distributivity, so the same identities hold. For example, with (a=-5,b=8), ((a+1)(b-1)-ab=(-4)(7)-(-40)=12), which also equals (b-a-1=8-(-5)-1=12).

((a+m)(b+n)=ab+an+bm+mn). Therefore the change from (ab) is (an+bm+mn). This signed expression also covers decreases when (m) or (n) is negative.

(i) Set (m=-2,n=3): ((a-2)(b+3)=ab+3a-2b-6), so the change is (3a-2b-6). (ii) Set (m=-3,n=-4): ((a-3)(b-4)=ab-4a-3b+12), so the change is (-4a-3b+12). Direct expansion gives the same results.

(i) ((a-u)(b+v)=ab+av-ub-uv). (ii) ((a-u)(b-v)=ab-av-ub+uv). Each result is obtained by multiplying every term in the first bracket by every term in the second.

Distributing (\\frac{3a}{2}) gives (\\frac32a^2-\\frac32ab+\\frac3{10}a).

No. The terms contain different letter parts, so none are like terms. The expression is already in its simplest collected form.

((a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2).

Distributing and collecting like terms gives (a^3+a^2b+2a^2b+2ab^2+ab^2+b^3=a^3+3a^2b+3ab^2+b^3).

The rows are ((p-1)(q-1),(p-1)q,(p-1)(q+1)); (p(q-1),pq,p(q+1)); and ((p+1)(q-1),(p+1)q,(p+1)(q+1)).

(i) (uv+3v-3u-9); (ii) (10+4a); (iii) (100ac+10ad+10bc+bd); (iv) (-x^2+9x-18); (v) (-5ac-5ad+bc+bd); (vi) (yz+5y+9z+45).

We need ((a+2)(b-4)=ab), which simplifies to (b=2a+4). Hence examples are (1\times6=3\times2=6), (2\times8=4\times4=16), and (3\times10=5\times6=30).

(i) (4a+5ab+ab^2-12b^2-3b^3). (ii) (4y^2+44yz-5y+77z-21).

The products are (a^2-b^2), (a^3-b^3), and (a^4-b^4). The next identity is ((a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5).

(3874\times11=42614) and (5678\times11=62458). Multiply by 10 and add the original number. Digitwise, keep the end digits and add adjacent digits from right to left, carrying whenever a sum is at least 10.

The products are (1034, 5445, 36069,) and (52703816), respectively.

(3874\times101=3874\times(100+1)=387400+3874=391274). Multiplying by 101 shifts the number two places left and adds the original number.

For (10^k+1), shift by (k) places and add; for (10^k-1), shift and subtract. Thus: (i) 8989; (ii) 95849; (iii) 266096831; (iv) 1112111; (v) 963666; (vi) 23454522.

The parts have areas (60^2=3600), (60\times5=300), (5\times60=300), and (5^2=25). Their sum is (4225), so (65^2=4225).

((30+35)^2=900+2100+1225=4225). Also, ((52+13)^2=2704+1352+169=4225). Both decompositions give the same area.

Their difference is (2ab). Therefore ((a+b)^2>a^2+b^2) when (ab>0), they are equal when (ab=0), and the left side is smaller when (ab<0).

(104^2=(100+4)^2=10000+800+16=10816). Also, (37^2=(40-3)^2=1600-240+9=1369).

The results are (m^2+6m+9), (p^2+12p+36), (36x^2+60x+25), and (9j^2+12jk+4k^2), respectively.

Subtracting the two (60\times5) strips removes their overlapping (5\times5) square twice, so add it back once: (60^2-2(60\times5)+5^2=3600-600+25=3025).

From a square of side (a), remove two (a\times b) strips and add back the (b\times b) overlap. Hence ((a-b)^2=a^2-2ab+b^2).

(99^2=(100-1)^2=10000-200+1=9801). (58^2=(60-2)^2=3600-240+4=3364).

The expansions are (b^2-12b+36), (4a^2-12a+9), and (49y^2-\frac{21}{2}yz+\frac9{16}z^2).

For any numbers (a,b), add the two square identities: ((a+b)^2+(a-b)^2=2a^2+2b^2=2(a^2+b^2)). Thus the two required squares are ((a+b)^2) and ((a-b)^2).

The identity is ((a+b)(a-b)=a^2-b^2). Hence (98\times102=(100-2)(100+2)=10000-4=9996), and (45\times55=(50-5)(50+5)=2500-25=2475).

Start with an (a\times a) square and remove a (b\times b) square. Rearrange the remaining L-shaped region into a rectangle with sides (a+b) and (a-b). Equal areas give (a^2-b^2=(a+b)(a-b)).

It is Identity 1C rearranged: ((a+b)(a-b)=a^2-b^2). Adding (b^2) to both sides gives (a^2=(a+b)(a-b)+b^2).

They are equal because (b-a=-(a-b)), and squaring removes the sign: ((b-a)^2=(a-b)^2).

One answer is (26^2-24^2=(26+24)(26-24)=50\times2=100).

The values are (164836, 5184, 21025, 1203409,) and (15376), respectively.

Yes. Their proofs use only addition, subtraction, multiplication, and distributivity, so they are polynomial identities valid for integers, rational numbers, and real numbers, not only counting numbers.

Correct results: (1) (15p^2-6pq); (2) (5x+10); (3) (3y+4); (4) (25m^2+60mn+36n^2); (5) (q^2-4q+4) is correct; (6) (18abc); (7) (5s) is correct; (8) (5w^2+6w) cannot be combined; (9) (5a^3+6a^2b+6ab^2); (10) (ab+4a+2b+8); (11) (ab(a+b+ab)) is correct; (12) (x^2+7x+10) is correct.

All four shown methods simplify to (k^2+2k=k(k+2)). The next figure, Step 4, has (4^2+2(4)=24) circles. Step 10 has (10^2+20=120) circles.

(15^2+2(15)=225+30=255) circles.

The first counts are 8, 12, 16, so Step (n) has ((n+2)^2-n^2=4n+4) tiles. Step 4 has 20 tiles and Step 10 has 44 tiles. Counting the four arms gives the same formula.

Subtracting the four rectangles gives ((m+n)^2-4mn=m^2-2mn+n^2=(n-m)^2). Thus the shaded square has area ((n-m)^2).

The expressions are (x^2-xy), (x(x+2y)-3xy), and (x(x-y)). Each simplifies to (x^2-xy=x(x-y)). At (x=8,y=3), the area is (8(8-3)=40) square units.

Subtracting the two overlapping strips gives (ps-pr-sr+r^2). Factoring gives the equivalent expression ((p-r)(s-r)). Substitution gives ((6-3.5)(9-3.5)=2.5\times5.5=13.75) square units.

(i) (46^2=(40+6)^2=2116). (ii) (397\times403=(400-3)(400+3)=159991). (iii) (91^2=(100-9)^2=8281). (iv) (43\times45=(44-1)(44+1)=1935).

(i) (p^2+10p-11); (ii) (9a^2-81b^2); (iii) (-6y^2-23y-20); (iv) (36x^2+60xy+25y^2); (v) (4x^2-2x+\frac14); (vi) (21p^2r+42pr).

(i) Two more than a square number is (s^2+2). (ii) The sum of squares of two consecutive numbers can be (m^2+(m+1)^2) or (m^2+(m-1)^2). The option (m^2+(m+1)^2) uses (m) as the smaller number.

For entries (a,a+1,a+7,a+8), the diagonal products differ by ((a+1)(a+7)-a(a+8)=a^2+8a+7-(a^2+8a)=7). The top-right to bottom-left product is always 7 larger.

(i) False: it simplifies to (k^2+2k-1). (ii) False: it is (4q^2-4q-3), which leaves remainder 1 on division by 4. (iii) True: ((2m)^2=4m^2), while ((2m+1)^2=8\frac{m(m+1)}2+1). (iv) False; for (n=1) the difference is 15, which is not 5 less than a square.

Modulo 7, the sum is (3+5=8\equiv1), and the product is (15\equiv1). The second number minus the first leaves (5-3=2); the first minus the second leaves (3-5\equiv5).

Let the numbers be (n-1,n,n+1). Then (n^2-(n-1)(n+1)=n^2-(n^2-1)=1). The result is always 1.

For numbers (a,b), the expression is ((a+b)\cdot\frac{a+b}{2}=\frac12(a+b)^2), exactly half the square of their sum.

(i) (16\times24=(20-4)(20+4)=400-16=384), while (14\times26=400-36=364); (16\times24) is larger. (ii) (26\times74=(50-24)(50+24)=2500-576=1924), while (25\times75=1875); (26\times74) is larger.

The outer rectangle has dimensions (2g+2w) and (g+2w). Subtracting the two green squares gives ((2g+2w)(g+2w)-2g^2=8w(g+w)) square feet.

(a) The counts follow ((y+2)^2): Step 4 has 36 and Step 10 has 144 basic units. (b) The counts follow ((y+1)^2+y=y^2+3y+1): Step 4 has 29 and Step 10 has 131 basic units.

The 10-coin triangle can be inverted in 3 moves, as stated. The 15-coin triangle requires a minimum of 5 moves: align the inverted triangle to maximise the coins that remain in place, then move only the non-overlapping coins. For larger triangles, the same maximum-overlap construction gives the minimum; draw the two orientations on the triangular lattice and count the unmatched positions.