Key Concepts & Formulae at a Glance
- An algebraic identity is an equation that is true for all values of the variables in it, unlike an ordinary equation, which need not hold for all values.
- Identities can be visualised geometrically using areas of squares/rectangles (2D) and volumes of cubes/cuboids (3D), and using algebra tiles.
- Identities can be used to factorise expressions, simplify rational expressions, and speed up numerical calculations like squares, cubes and products.
I. \((a+b)^2 = a^2+2ab+b^2\)
II. \((a-b)^2 = a^2-2ab+b^2\)
III. \((a+b)(a-b) = a^2-b^2\)
IV. \((x+a)(x+b) = x^2+(a+b)x+ab\)
V. \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\)
VI. \((a+b)^3 = a^3+b^3+3ab(a+b)\)
VII. \((a-b)^3 = a^3-b^3-3ab(a-b)\)
VIII. \(a^3+b^3 = (a+b)(a^2-ab+b^2)\)
IX. \(a^3-b^3 = (a-b)(a^2+ab+b^2)\)
X. \(a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
Also used in this chapter: \((ax+b)(cx+d) = acx^2+(ad+bc)x+bd\)
Think and Reflect
TRTry and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Let us take four consecutive integers \(n, n+1, n+2, n+3\), so their squares are \(n^2, (n+1)^2, (n+2)^2, (n+3)^2\).
Try adding the first and the last squares, and subtracting the sum of the two middle squares:
\(n^2+(n+3)^2-\left[(n+1)^2+(n+2)^2\right]\)
\(=n^2+n^2+6n+9-\left(n^2+2n+1\right)-\left(n^2+4n+4\right)\)
\(=2n^2+6n+9-n^2-2n-1-n^2-4n-4\)
\(=(2n^2-n^2-n^2)+(6n-2n-4n)+(9-1-4)=0+0+4=4\)
Check with actual numbers: 1, 4, 9, 16 → \(1+16-(4+9)=17-13=4\). Also 4, 9, 16, 25 → \(4+25-(9+16)=29-25=4\). The result is always 4, no matter which four consecutive squares we pick.
Think and Reflect
TR1. What can you say about a and b if \((a+b)^2 < a^2+b^2\)?
2. What can you say about a and b if \((a+b)^2 > a^2+b^2\)?
3. When will \((a+b)^2\) be equal to \(a^2+b^2\)?
We know that \((a+b)^2=a^2+2ab+b^2\), so \((a+b)^2-(a^2+b^2)=2ab\).
1. \((a+b)^2 < a^2+b^2\) exactly when \(2ab < 0\), i.e. when \(ab < 0\) — this happens when a and b have opposite signs (one positive, one negative).
2. \((a+b)^2 > a^2+b^2\) exactly when \(2ab > 0\), i.e. when \(ab > 0\) — this happens when a and b have the same sign (both positive or both negative).
3. \((a+b)^2 = a^2+b^2\) exactly when \(2ab=0\), i.e. when \(ab=0\) — this happens when at least one of a, b is 0.
Exercise Set 4.1
1Using the identity \((a+b)^2=a^2+2ab+b^2\), expand the following:
(i) \((7x+4y)^2\)
(ii) \(\left(\frac{7}{5}x+\frac{3}{2}y\right)^2\)
(iii) \((2.5p+1.5q)^2\)
(iv) \(\left(\frac{3}{4}s+8t\right)^2\)
(v) \(\left(x+\frac{1}{2y}\right)^2\)
(vi) \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
In each part, we match the expression to \((a+b)^2=a^2+2ab+b^2\) and substitute.
(i) Here \(a=7x, b=4y\): \((7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2=49x^2+56xy+16y^2\)
(ii) Here \(a=\frac{7}{5}x, b=\frac{3}{2}y\): \(\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\left(\frac{7}{5}x\right)^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+\left(\frac{3}{2}y\right)^2=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2\)
(iii) Here \(a=2.5p, b=1.5q\): \((2.5p+1.5q)^2=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2=6.25p^2+7.5pq+2.25q^2\)
(iv) Here \(a=\frac{3}{4}s, b=8t\): \(\left(\frac{3}{4}s+8t\right)^2=\left(\frac{3}{4}s\right)^2+2\left(\frac{3}{4}s\right)(8t)+(8t)^2=\frac{9}{16}s^2+12st+64t^2\)
(v) Here \(a=x, b=\frac{1}{2y}\): \(\left(x+\frac{1}{2y}\right)^2=x^2+2(x)\left(\frac{1}{2y}\right)+\left(\frac{1}{2y}\right)^2=x^2+\frac{x}{y}+\frac{1}{4y^2}\)
(vi) Here \(a=\frac{1}{x}, b=\frac{1}{y}\): \(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}\)
2Using the same identity, find the values of the following:
(i) \((64)^2\)
(ii) \((105)^2\)
(iii) \((205)^2\)
We write each number as a convenient sum \(a+b\) and apply \((a+b)^2=a^2+2ab+b^2\).
(i) \(64=60+4\): \((64)^2=(60+4)^2=60^2+2(60)(4)+4^2=3600+480+16=4096\)
(ii) \(105=100+5\): \((105)^2=(100+5)^2=100^2+2(100)(5)+5^2=10000+1000+25=11025\)
(iii) \(205=200+5\): \((205)^2=(200+5)^2=200^2+2(200)(5)+5^2=40000+2000+25=42025\)
Think and Reflect
TRWhat if we replace b by −b in \((a+b)^2=a^2+2ab+b^2\)?
Replacing b by −b: \(\left[a+(-b)\right]^2=a^2+2a(-b)+(-b)^2\)
\((a-b)^2=a^2-2ab+b^2\)
Exercise Set 4.2
1Factor completely:
(i) \(9x^2+24xy+16y^2\)
(ii) \(4s^2+20st+25t^2\)
(iii) \(49x^2+28xy+4y^2\)
(iv) \(64p^2+\frac{32}{3}pq+\frac{4}{9}q^2\)
(v)* \(3a^2+4ab+\frac{4}{3}b^2\)
(vi)* \(\frac{9}{5}s^2+6sv+5v^2\)
We compare each expression with \(a^2+2ab+b^2=(a+b)^2\). For (v) and (vi), we first take a common factor out, as in Example 7, so that what remains is a perfect square.
(i) \(9x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2=(3x+4y)^2\)
(ii) \(4s^2+20st+25t^2=(2s)^2+2(2s)(5t)+(5t)^2=(2s+5t)^2\)
(iii) \(49x^2+28xy+4y^2=(7x)^2+2(7x)(2y)+(2y)^2=(7x+2y)^2\)
(iv) \(64p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p)^2+2(8p)\left(\frac{2}{3}q\right)+\left(\frac{2}{3}q\right)^2=\left(8p+\frac{2}{3}q\right)^2\)
(v) Take out \(\frac{1}{3}\) as a common factor: \(3a^2+4ab+\frac{4}{3}b^2=\frac{1}{3}\left(9a^2+12ab+4b^2\right)=\frac{1}{3}\left[(3a)^2+2(3a)(2b)+(2b)^2\right]=\frac{1}{3}(3a+2b)^2\)
(vi) Take out \(\frac{1}{5}\) as a common factor: \(\frac{9}{5}s^2+6sv+5v^2=\frac{1}{5}\left(9s^2+30sv+25v^2\right)=\frac{1}{5}\left[(3s)^2+2(3s)(5v)+(5v)^2\right]=\frac{1}{5}(3s+5v)^2\)
2Find the values of the following using the identity \((a-b)^2=a^2-2ab+b^2\):
(i) \((79)^2\)
(ii) \((193)^2\)
(iii) \((299)^2\)
We write each number as a convenient difference \(a-b\) and apply \((a-b)^2=a^2-2ab+b^2\).
(i) \(79=80-1\): \((79)^2=(80-1)^2=80^2-2(80)(1)+1^2=6400-160+1=6241\)
(ii) \(193=200-7\): \((193)^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249\)
(iii) \(299=300-1\): \((299)^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401\)
Think and Reflect
TRLabel the squares and rectangles in Fig. 4.4 so that it represents the identity \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\).
Fig. 4.4 labelled: a 3×3 grid, strip widths a, b, c along each side
Fig. 4.4 is a big square of side \(a+b+c\), split by two horizontal and two vertical lines into a 3×3 grid of 9 smaller regions, with the top-to-bottom and left-to-right strip widths being a, b, c in that order.
The three squares along the diagonal have areas \(a^2\), \(b^2\), \(c^2\) (dimensions \(a\times a\), \(b\times b\), \(c\times c\)).
Each pair of variables gives two equal rectangles: two rectangles of area ab (dimensions \(a\times b\)), two of area bc (dimensions \(b\times c\)), and two of area ca (dimensions \(c\times a\)).
Adding all 9 regions: \(a^2+b^2+c^2+ab+ab+bc+bc+ca+ca=a^2+b^2+c^2+2ab+2bc+2ca\), which is the area of the whole square, \((a+b+c)^2\).
TR1. Try to evaluate 35², 65², 85², 105² using a suitable identity. Do you observe any interesting pattern?
2. Observe the two rows of figures in Fig. 4.6. They represent an algebraic identity. Try to identify it.
Part 1: Every one of these numbers ends in 5, so we can write it as \(10k+5\) and use \((a+b)^2=a^2+2ab+b^2\) with \(a=10k, b=5\):
\((10k+5)^2=(10k)^2+2(10k)(5)+5^2=100k^2+100k+25=100k(k+1)+25\)
\(35=10(3)+5\): \(35^2=100(3)(4)+25=1200+25=1225\)
\(65=10(6)+5\): \(65^2=100(6)(7)+25=4200+25=4225\)
\(85=10(8)+5\): \(85^2=100(8)(9)+25=7200+25=7225\)
\(105=10(10)+5\): \(105^2=100(10)(11)+25=11000+25=11025\)
Pattern: for any number ending in 5, written as \(10k+5\), its square always ends in 25, and the digits before the 25 are exactly \(k(k+1)\) — that is, k multiplied by the next whole number.
Part 2: Fig. 4.6 shows a square of side \(2a+2b+2c\) built from four smaller squares of sides \((a+b+c)\), \((a+b-c)\), \((a-b+c)\), \((a-b-c)\), alongside three separate squares of sides \(2a\), \(2b\), \(2c\). Expanding all four squares and adding:
Expanding all four squares and adding:
\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\) …(1)
\((a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca\) …(2)
\((a-b+c)^2=a^2+b^2+c^2-2ab-2bc+2ca\) …(3)
\((a-b-c)^2=a^2+b^2+c^2-2ab+2bc-2ca\) …(4)
Adding (1), (2), (3) and (4): every \(ab\), \(bc\), \(ca\) term cancels in pairs, leaving \(4a^2+4b^2+4c^2\), which is exactly \((2a)^2+(2b)^2+(2c)^2\).
Fig. 4.6 (illustrative): four squares of sides (a+b+c), (a+b−c), (a−b+c), (a−b−c) have the same total area as squares of sides 2a, 2b, 2c
Expanding all four squares and adding:
\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\) …(1)
\((a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca\) …(2)
\((a-b+c)^2=a^2+b^2+c^2-2ab-2bc+2ca\) …(3)
\((a-b-c)^2=a^2+b^2+c^2-2ab+2bc-2ca\) …(4)
Adding (1), (2), (3) and (4): every \(ab\), \(bc\), \(ca\) term cancels in pairs, leaving \(4a^2+4b^2+4c^2\), which is exactly \((2a)^2+(2b)^2+(2c)^2\).
Exercise Set 4.3
1Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier:
(i) \(117^2\)
(ii) \(78^2\)
(iii) \(198^2\)
(iv) \(214^2\)
(v) \(1104^2\)
(vi) \(1120^2\)
(i) \(117=100+17\), use \((a+b)^2\): \(117^2=100^2+2(100)(17)+17^2=10000+3400+289=13689\)
(ii) \(78=80-2\), use \((a-b)^2\): \(78^2=80^2-2(80)(2)+2^2=6400-320+4=6084\)
(iii) \(198=200-2\), use \((a-b)^2\): \(198^2=200^2-2(200)(2)+2^2=40000-800+4=39204\)
(iv) \(214=200+14\), use \((a+b)^2\): \(214^2=200^2+2(200)(14)+14^2=40000+5600+196=45796\)
(v) \(1104=1100+4\), use \((a+b)^2\): \(1104^2=1100^2+2(1100)(4)+4^2=1210000+8800+16=1218816\)
(vi) \(1120=1100+20\), use \((a+b)^2\): \(1120^2=1100^2+2(1100)(20)+20^2=1210000+44000+400=1254400\)
2Factor using suitable identities:
(i) \(16y^2-24y+9\)
(ii) \(\frac{9}{4}s^2+6st+4t^2\)
(iii) \(\frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+3nk+2mn+9n^2\)
(iv) \(\frac{p^2}{16}-2+\frac{16}{p^2}\)
(v) \(9a^2+4b^2+c^2-12ab+6ac-4bc\)
(i) \(16y^2-24y+9=(4y)^2-2(4y)(3)+3^2=(4y-3)^2\)
(ii) \(\frac{9}{4}s^2+6st+4t^2=\left(\frac{3}{2}s\right)^2+2\left(\frac{3}{2}s\right)(2t)+(2t)^2=\left(\frac{3}{2}s+2t\right)^2\)
(iii) This has three variables, so try \((a+b+c)^2\) with \(a=\dfrac{m}{3}, b=\dfrac{k}{2}, c=3n\).
Check: \(a^2=\dfrac{m^2}{9}\), \(b^2=\dfrac{k^2}{4}\), \(c^2=9n^2\)
\(2ab=2\left(\dfrac{m}{3}\right)\left(\dfrac{k}{2}\right)=\dfrac{mk}{3}\)
\(2bc=2\left(\dfrac{k}{2}\right)(3n)=3nk\)
\(2ac=2\left(\dfrac{m}{3}\right)(3n)=2mn\)
All terms match, so \(\dfrac{m^2}{9}+\dfrac{mk}{3}+\dfrac{k^2}{4}+3nk+2mn+9n^2=\left(\dfrac{m}{3}+\dfrac{k}{2}+3n\right)^2\)
(iv) Here \(a=\dfrac{p}{4}, b=\dfrac{4}{p}\).
\(a^2=\dfrac{p^2}{16}\), \(b^2=\dfrac{16}{p^2}\), \(2ab=2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right)=2\)
So \(\dfrac{p^2}{16}-2+\dfrac{16}{p^2}=\left(\dfrac{p}{4}\right)^2-2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right)+\left(\dfrac{4}{p}\right)^2=\left(\dfrac{p}{4}-\dfrac{4}{p}\right)^2\)
(v) Try \((3a-2b+c)^2\) using \((x+y+z)^2\) with \(x=3a, y=-2b, z=c\).
\(x^2+y^2+z^2=9a^2+4b^2+c^2\)
\(2xy=2(3a)(-2b)=-12ab\)
\(2yz=2(-2b)(c)=-4bc\)
\(2zx=2(c)(3a)=6ac\)
All terms match, so \(9a^2+4b^2+c^2-12ab+6ac-4bc=(3a-2b+c)^2\)
3Expand the following using the identity \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\):
(i) \((p+3q+7r)^2\)
(ii) \((3x-2y+4z)^2\)
(i) With \(a=p, b=3q, c=7r\): \((p+3q+7r)^2=p^2+(3q)^2+(7r)^2+2(p)(3q)+2(3q)(7r)+2(7r)(p)\)
\(=p^2+9q^2+49r^2+6pq+42qr+14rp\)
(ii) With \(a=3x, b=-2y, c=4z\): \((3x-2y+4z)^2=(3x)^2+(-2y)^2+(4z)^2+2(3x)(-2y)+2(-2y)(4z)+2(4z)(3x)\)
\(=9x^2+4y^2+16z^2-12xy-16yz+24xz\)
4Is this an identity? \((a+b-c)^2+(a-b+c)^2+(a-b-c)^2=2a^2+2b^2+2c^2\)
We expand each square using \((x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\).
\((a+b-c)^2\): here \(x=a,y=b,z=-c\), giving \(a^2+b^2+c^2+2ab-2bc-2ac\) …(1)
\((a-b+c)^2\): here \(x=a,y=-b,z=c\), giving \(a^2+b^2+c^2-2ab-2bc+2ac\) …(2)
\((a-b-c)^2\): here \(x=a,y=-b,z=-c\), giving \(a^2+b^2+c^2-2ab+2bc-2ac\) …(3)
Adding (1), (2) and (3): \(3a^2+3b^2+3c^2+(2ab-2ab-2ab)+(-2bc-2bc+2bc)+(-2ac+2ac-2ac)\)
\(=3a^2+3b^2+3c^2-2ab-2bc-2ac\)
This is not the same as \(2a^2+2b^2+2c^2\) for all values of a, b, c. For example, take \(a=1, b=0, c=0\): LHS \(=1^2+1^2+1^2=3\), but RHS \(=2(1)^2+0+0=2\). Since \(3\neq 2\), the two sides are not equal for all values.
Think and Reflect
TRSuppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
For a split \(ax+bx\) of \(7x\) to form a rectangle for \(x^2+7x+12\), we need \(a+b=7\) (the x-term) and \(ab=12\) (the constant term, since the corner needs exactly \(ab\) unit tiles).
With the split \(2x+5x\): \(a=2, b=5\).
\(a+b=2+5=7\) ✓, but \(ab=2\times5=10\neq12\) ✗
So this split does not form a valid rectangle — only 10 unit tiles fit the corner, and the remaining 2 tiles cannot complete a clean rectangle.
Checking other splits of 7: only \(a=3, b=4\) gives \(ab=3\times4=12\), matching the 12 unit tiles needed.
This confirms \((x+3)(x+4)\) is correct; \((x+2)(x+5)\) would instead give constant term \(2\times5=10\), i.e. \(x^2+7x+10\) — a different expression.
TRAlgebra tiles can be used to represent products and find factors.
1. Figure out the product of x + 2 and x + 3 using algebra tiles.
2. Lay out algebra tiles for \(x^2+11x+30\) in such a way that you will see its factors.
Part 1: (x+2)(x+3) tiled — 1 x², 5 x-tiles, 6 unit tiles
Part 1: Represent \(x+2\) using one x-tile and 2 unit tiles, and \(x+3\) using one x-tile and 3 unit tiles, as the two sides of a rectangle. The rectangle then contains one \(x^2\)-tile, 2 x-tiles along one side, 3 x-tiles along the other, and \(2\times3=6\) unit tiles arranged in a 2-by-3 grid.
\((x+2)(x+3)=x^2+2x+3x+6=x^2+5x+6\)
Part 2: For \(x^2+11x+30\), we need two numbers that add to 11 and multiply to 30 — these are 5 and 6 (since \(5+6=11\) and \(5\times6=30\)). Arrange one \(x^2\)-tile, 5 x-tiles along one side, 6 x-tiles along the other side, and the 30 unit tiles in a matching 5-by-6 grid in the corner. The resulting rectangle has dimensions \((x+5)\) by \((x+6)\).
Part 2: x²+11x+30 tiled as (x+5)(x+6)
Part 2: For \(x^2+11x+30\), we need two numbers that add to 11 and multiply to 30 — these are 5 and 6 (since \(5+6=11\) and \(5\times6=30\)). Arrange one \(x^2\)-tile, 5 x-tiles along one side, 6 x-tiles along the other side, and the 30 unit tiles in a matching 5-by-6 grid in the corner. The resulting rectangle has dimensions \((x+5)\) by \((x+6)\).
TRWe have seen that \((x+3)(x+4)=x^2+7x+12\). Also \((x+6)(x+7)=x^2+13x+42\). Generalise the pattern to get an expression for \((x+a)(x+b)\).
In both examples, the coefficient of x in the result equals the sum of the two constants, and the constant term equals their product: for \((x+3)(x+4)\), \(3+4=7\) and \(3\times4=12\); for \((x+6)(x+7)\), \(6+7=13\) and \(6\times7=42\).
Generalising, using the distributive property: \((x+a)(x+b)=x(x+b)+a(x+b)=x^2+bx+ax+ab=x^2+(a+b)x+ab\)
TRConsider a rectangle of side-lengths \(2x+3\) and \(3x+1\) (Fig. 4.8). Fill in the blanks: \((px+a)(qx+b)=(\;\;)x^2+(\;\;)x+\;\;\).
Fig. 4.8: algebra tiles for (2x+3)(3x+1) — 6 x²-tiles, 11 x-tiles, 3 unit tiles
\((px+a)(qx+b)=px(qx+b)+a(qx+b)=pqx^2+pbx+aqx+ab=pqx^2+(pb+aq)x+ab\)
Check with \((2x+3)(3x+1)\): \(p=2,a=3,q=3,b=1\), so \(pq=6\), \(pb+aq=2(1)+3(3)=11\), \(ab=3\) — matching direct expansion \(6x^2+11x+3\).
Think and Reflect
TRJames and Reshma were talking about algebraic identities. James: \((a-b)^2(a+b)=(a^2-2ab+b^2)(a+b)\). Reshma: I have a different idea. \((a-b)^2(a+b)=(a-b)[(a-b)(a+b)]=(a-b)(a^2-b^2)\). I will find this product to get the answer. According to you, who is correct and why?
Both James and Reshma are correct — they are just grouping the three factors \((a-b)\), \((a-b)\), and \((a+b)\) differently, and multiplication of algebraic expressions can be grouped in any order (it is associative and commutative), so both paths must lead to the same final answer.
James's way: \((a^2-2ab+b^2)(a+b)=a^3+a^2b-2a^2b-2ab^2+ab^2+b^3=a^3-a^2b-ab^2+b^3\)
Reshma's way: \((a-b)(a^2-b^2)=a^3-ab^2-a^2b+b^3=a^3-a^2b-ab^2+b^3\)
Both give the same final expression, \(a^3-a^2b-ab^2+b^3\), confirming that either grouping works — Reshma's route is a little quicker here because \((a-b)(a+b)=a^2-b^2\) is a familiar identity that simplifies the next step.
Exercise Set 4.4
1Fill in the blanks to complete the following identities:
(i) \(s^2-11s+24=(\;\;)(\;\;)\)
(ii) \((\;\;)(x+1)=(3x^2-4x-7)\)
(iii) \(10x^2-11x-6=(2x-\;\;)(\;\;+2)\)
(iv) \(6x^2+7x+2=(\;\;)(\;\;)\)
(i) We need \(a+b=-11\) and \(ab=24\). Both numbers must be negative since their sum is negative but product is positive: \(a=-3, b=-8\) works, since \(-3+(-8)=-11\) and \((-3)(-8)=24\). So \(s^2-11s+24=(s-3)(s-8)\)
(ii) Dividing \(3x^2-4x-7\) by \((x+1)\):
\(3x^2\div x=3x\); \(3x(x+1)=3x^2+3x\); subtracting leaves \(-7x-7\)
\(-7x\div x=-7\); \(-7(x+1)=-7x-7\); subtracting leaves 0
So the quotient is \(3x-7\), and \((3x-7)(x+1)=3x^2-4x-7\)
(iii) We need to factor \(10x^2-11x-6\). Trying \((2x-3)(5x+2)\): \((2x-3)(5x+2)=10x^2+4x-15x-6=10x^2-11x-6\) ✓. So the blanks are \((2x-3)(5x+2)\)
(iv) For \(6x^2+7x+2\), we need two numbers multiplying to \(6\times2=12\) and adding to 7: these are 3 and 4. Split the middle term: \(6x^2+3x+4x+2=3x(2x+1)+2(2x+1)=(3x+2)(2x+1)\)
2Select and use the identity that will help you to find the following products without multiplying directly:
(i) \((41)^2\)
(ii) \((27)^2\)
(iii) \((23\times17)\)
(iv) \((135)^2\)
(v) \((97)^2\)
(vi) \((18\times29)\)
(vii) \((34\times43)\)
(viii) \((205)^2\)
(i) \(41=40+1\): \(41^2=(40+1)^2=1600+80+1=1681\)
(ii) \(27=30-3\): \(27^2=(30-3)^2=900-180+9=729\)
(iii) \(23=20+3, 17=20-3\): \(23\times17=(20+3)(20-3)=20^2-3^2=400-9=391\)
(iv) \(135=130+5\): \(135^2=(130+5)^2=16900+1300+25=18225\)
(v) \(97=100-3\): \(97^2=(100-3)^2=10000-600+9=9409\)
(vi) For \(18\times29\), take the average \(\frac{18+29}{2}=23.5\) and the half-difference \(\frac{29-18}{2}=5.5\), so \(18=23.5-5.5\) and \(29=23.5+5.5\): \(18\times29=(23.5-5.5)(23.5+5.5)=23.5^2-5.5^2=552.25-30.25=522\)
(vii) Similarly for \(34\times43\): average \(=\frac{34+43}{2}=38.5\), half-difference \(=\frac{43-34}{2}=4.5\): \(34\times43=(38.5-4.5)(38.5+4.5)=38.5^2-4.5^2=1482.25-20.25=1462\)
(viii) \(205=200+5\): \(205^2=(200+5)^2=40000+2000+25=42025\)
3Factor the following:
(i) \(9a^2+b^2+4c^2-6ab+12ac-4bc\)
(ii) \(16s^2+25t^2-40st\)
(iii) \(r^2-r-42\)
(iv) \(49g^2+14gh+h^2\)
(v) \(64u^2+121v^2+4w^2-176uv-32uw+44vw\)
(i) Try \((3a-b+2c)^2\) using \(x=3a,y=-b,z=2c\).
\(x^2+y^2+z^2=9a^2+b^2+4c^2\)
\(2xy=-6ab\), \(2yz=-4bc\), \(2zx=12ac\) — all match
So \(9a^2+b^2+4c^2-6ab-4bc+12ac=(3a-b+2c)^2\)
(ii) \(16s^2+25t^2-40st=(4s)^2-2(4s)(5t)+(5t)^2=(4s-5t)^2\)
(iii) We need \(a+b=-1\) and \(ab=-42\): these are \(a=-7,b=6\). So \(r^2-r-42=(r-7)(r+6)\)
(iv) \(49g^2+14gh+h^2=(7g)^2+2(7g)(h)+h^2=(7g+h)^2\)
(v) Try \((8u-11v-2w)^2\) using \(x=8u,y=-11v,z=-2w\).
\(x^2+y^2+z^2=64u^2+121v^2+4w^2\)
\(2xy=2(8u)(-11v)=-176uv\)
\(2yz=2(-11v)(-2w)=44vw\)
\(2zx=2(-2w)(8u)=-32uw\) — all match
So \(64u^2+121v^2+4w^2-176uv+44vw-32uw=(8u-11v-2w)^2\)
Think and Reflect
TRWe already know that \(x^2-y^2=(x-y)(x+y)\). Further, we have verified that \(x^3-y^3=(x-y)(x^2+xy+y^2)\). Observe that \(x-y\) is a common factor of \(x^2-y^2\) and \(x^3-y^3\). Do you think \(x-y\) is also a factor of \(x^4-y^4\)? Note that \(x^4-y^4=(x^2)^2-(y^2)^2=(x^2-y^2)(x^2+y^2)\). Can you see how \(x-y\) is a factor of \(x^4-y^4\)? How about \(x^5-y^5\)? Does this also have \(x-y\) as a factor?
Yes. Since \(x^4-y^4=(x^2-y^2)(x^2+y^2)\), and we already know \(x^2-y^2=(x-y)(x+y)\), we can substitute:
\(x^4-y^4=(x-y)(x+y)(x^2+y^2)\)
This clearly shows \((x-y)\) as one of the factors.
For \(x^5-y^5\), the same pattern continues:
\(x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)\)
This can be checked by expanding the right-hand side using the distributive property — all the middle terms cancel in pairs, leaving just \(x^5-y^5\).
So \((x-y)\) is a factor here too.
TRPredict what \((x+y)(x^2-xy+y^2)\) will be.
Following the same pattern as \((x-y)(x^2+xy+y^2)=x^3-y^3\), but with a plus sign, we expect the result to be \(x^3+y^3\). Let's verify using the distributive property:
\((x+y)(x^2-xy+y^2)=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)
\(=x^3+(-x^2y+x^2y)+(xy^2-xy^2)+y^3=x^3+y^3\)
Think and Reflect
TRTry to simplify the following rational expression (assuming \(t^2+2ts-48s^2\neq0\)): \(\dfrac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\dfrac{(6s-t)^2}{(\;\;+\;\;)(\;\;+\;\;)}\)
The numerator is already given as a perfect square: \(36s^2-12st+t^2=(6s)^2-2(6s)(t)+t^2=(6s-t)^2\).
For the denominator, treat it as a quadratic in t: we need two terms adding to \(2s\) and multiplying to \(-48s^2\) — these are \(8s\) and \(-6s\), since \(8s+(-6s)=2s\) and \((8s)(-6s)=-48s^2\).
\(t^2+2ts-48s^2=t^2+8st-6st-48s^2=t(t+8s)-6s(t+8s)=(t-6s)(t+8s)\)
So \(\dfrac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\dfrac{(6s-t)^2}{(t-6s)(t+8s)}\). Since \((6s-t)^2=(t-6s)^2\) (squaring removes the sign difference), one factor of \((t-6s)\) cancels with a factor of \((6s-t)^2=(t-6s)^2\):
\(=\dfrac{(t-6s)^2}{(t-6s)(t+8s)}=\dfrac{t-6s}{t+8s}\)
Exercise Set 4.5
1Simplify the following rational expressions, assuming that the expressions in the denominators are not equal to zero:
(i) \(\dfrac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}\)
(ii) \(\dfrac{n^3-3n^2m+3nm^2-m^3}{5m^2-10mn+5n^2}\)
(iii) \(\dfrac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}\)
(iv) \(\dfrac{4y^2-20yz+25z^2}{25z^2-4y^2}\)
(v) \(\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}\)
(vi) \(\dfrac{p^4-16}{p^2-4p+4}\)
(i) Numerator: \(3p^2-3pq-18q^2=3(p^2-pq-6q^2)=3(p+2q)(p-3q)\) [since \(2q+(-3q)=-q\) and \((2q)(-3q)=-6q^2\)]
Denominator: \(p^2+3pq-10q^2=(p+5q)(p-2q)\) [since \(5q+(-2q)=3q\) and \((5q)(-2q)=-10q^2\)]
The numerator and denominator share no common factor, so the expression is already in its simplest form: \(\dfrac{3(p+2q)(p-3q)}{(p+5q)(p-2q)}\)
(ii) Numerator: \(n^3-3n^2m+3nm^2-m^3\) matches \((a-b)^3=a^3-3a^2b+3ab^2-b^3\) with \(a=n,b=m\), so it equals \((n-m)^3\).
Denominator: \(5m^2-10mn+5n^2=5(m^2-2mn+n^2)=5(m-n)^2=5(n-m)^2\)
\(\dfrac{(n-m)^3}{5(n-m)^2}=\dfrac{n-m}{5}\)
(iii) The numerator \(w^3-v^3+x^3+3wvx\) matches \(a^3+b^3+c^3-3abc\) with \(a=w, b=-v, c=x\), since \(-3abc=-3(w)(-v)(x)=3wvx\). So numerator \(=(w-v+x)\left(w^2+v^2+x^2+wv+vx-wx\right)\) using \(a^2+b^2+c^2-ab-bc-ca\).
The denominator \(w^2+v^2+x^2-2wv-2vx+2wx\) matches \((x+y+z)^2\) with \(x=w,y=-v,z=x\): \((w-v+x)^2=w^2+v^2+x^2-2wv-2vx+2wx\) ✓
\(\dfrac{(w-v+x)(w^2+v^2+x^2+wv+vx-wx)}{(w-v+x)^2}=\dfrac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}\)
(iv) Numerator: \(4y^2-20yz+25z^2=(2y)^2-2(2y)(5z)+(5z)^2=(2y-5z)^2=(5z-2y)^2\)
Denominator: \(25z^2-4y^2=(5z-2y)(5z+2y)\)
\(\dfrac{(5z-2y)^2}{(5z-2y)(5z+2y)}=\dfrac{5z-2y}{5z+2y}\)
(v) \(x^2+x-6=(x+3)(x-2)\); \(x^2-7x+12=(x-3)(x-4)\); \(x^2-6x+8=(x-2)(x-4)\); \(x^2-9=(x-3)(x+3)\)
Numerator \(=(x+3)(x-2)(x-3)(x-4)\); Denominator \(=(x-2)(x-4)(x-3)(x+3)\) — these are identical (just written in a different order), so the whole expression simplifies to 1.
(vi) Numerator: \(p^4-16=(p^2)^2-4^2=(p^2-4)(p^2+4)=(p-2)(p+2)(p^2+4)\)
Denominator: \(p^2-4p+4=(p-2)^2\)
\(\dfrac{(p-2)(p+2)(p^2+4)}{(p-2)^2}=\dfrac{(p+2)(p^2+4)}{p-2}\)
End-of-Chapter Exercises
1Use suitable identities to find the following products:
(i) \((-3x+4)^2\)
(ii) \((2s+7)(2s-7)\)
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)\)
(iv) \((2n+7)(2n-7)\)
(v) \((s-2t)(s^2+2st+4t^2)\)
(vi) \(\left(\frac{1}{2r}-4r\right)^2\)
(vii) \((-3m+4k-l)^2\)
(viii) \(\left(x-\frac{1}{3}y\right)^3\)
(ix) \(\left(\frac{7}{2}k-\frac{2}{3}m\right)^3\)
(i) \((-3x+4)^2=(4-3x)^2=4^2-2(4)(3x)+(3x)^2=16-24x+9x^2\)
(ii) \((2s+7)(2s-7)=(2s)^2-7^2=4s^2-49\)
(iii) \(\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)=(p^2)^2-\left(\frac{1}{2}\right)^2=p^4-\frac{1}{4}\)
(iv) \((2n+7)(2n-7)=(2n)^2-7^2=4n^2-49\)
(v) This matches \(a^3-b^3=(a-b)(a^2+ab+b^2)\) with \(a=s, b=2t\).
Check: \(a^2+ab+b^2=s^2+2st+4t^2\) exactly matches.
So \((s-2t)(s^2+2st+4t^2)=s^3-(2t)^3=s^3-8t^3\)
(vi) \(\left(\frac{1}{2r}-4r\right)^2=\left(\frac{1}{2r}\right)^2-2\left(\frac{1}{2r}\right)(4r)+(4r)^2=\frac{1}{4r^2}-4+16r^2\)
(vii) Using \((x+y+z)^2\) with \(x=-3m, y=4k, z=-l\).
\(x^2+y^2+z^2=9m^2+16k^2+l^2\)
\(2xy+2yz+2zx=2(-3m)(4k)+2(4k)(-l)+2(-l)(-3m)=-24mk-8kl+6lm\)
So \((-3m+4k-l)^2=9m^2+16k^2+l^2-24mk-8kl+6lm\)
(viii) Using \((a-b)^3=a^3-3a^2b+3ab^2-b^3\) with \(a=x, b=\dfrac{y}{3}\).
\(3a^2b=3x^2\cdot\dfrac{y}{3}=x^2y\)
\(3ab^2=3x\cdot\dfrac{y^2}{9}=\dfrac{xy^2}{3}\)
\(b^3=\dfrac{y^3}{27}\)
So \(\left(x-\dfrac{y}{3}\right)^3=x^3-x^2y+\dfrac{xy^2}{3}-\dfrac{y^3}{27}\)
(ix) Using \((a-b)^3\) with \(a=\dfrac{7k}{2}, b=\dfrac{2m}{3}\).
\(a^3=\dfrac{343k^3}{8}\)
\(3a^2b=3\cdot\dfrac{49k^2}{4}\cdot\dfrac{2m}{3}=\dfrac{49k^2m}{2}\)
\(3ab^2=3\cdot\dfrac{7k}{2}\cdot\dfrac{4m^2}{9}=\dfrac{14km^2}{3}\)
\(b^3=\dfrac{8m^3}{27}\)
So \(\left(\dfrac{7k}{2}-\dfrac{2m}{3}\right)^3=\dfrac{343k^3}{8}-\dfrac{49k^2m}{2}+\dfrac{14km^2}{3}-\dfrac{8m^3}{27}\)
2Find the values using suitable identities:
(i) \(17\times21\)
(ii) \(104\times96\)
(iii) \(24\times16\)
(iv) \(147^3\)
(v) \(199^3\)
(vi) \(127^3\)
(vii) \((-107)^3\)
(viii) \((-299)^3\)
(i) \(17=19-2, 21=19+2\): \(17\times21=(19-2)(19+2)=19^2-2^2=361-4=357\)
(ii) \(104=100+4, 96=100-4\): \(104\times96=(100+4)(100-4)=100^2-4^2=10000-16=9984\)
(iii) \(24=20+4, 16=20-4\): \(24\times16=(20+4)(20-4)=20^2-4^2=400-16=384\)
(iv) \(147=150-3\), using \((a-b)^3=a^3-3a^2b+3ab^2-b^3\):
\(147^3=150^3-3(150)^2(3)+3(150)(3)^2-3^3\)
\(=3375000-202500+4050-27=3176523\)
(v) \(199=200-1\):
\(199^3=200^3-3(200)^2(1)+3(200)(1)^2-1^3\)
\(=8000000-120000+600-1=7880599\)
(vi) \(127=130-3\):
\(127^3=130^3-3(130)^2(3)+3(130)(3)^2-3^3\)
\(=2197000-152100+3510-27=2048383\)
(vii) \((-107)^3=-(107)^3\).
Using \(107=100+7\): \(107^3=100^3+3(100)^2(7)+3(100)(7)^2+7^3\)
\(=1000000+210000+14700+343=1225043\)
So \((-107)^3=-1225043\)
(viii) \((-299)^3=-(299)^3\).
Using \(299=300-1\): \(299^3=300^3-3(300)^2(1)+3(300)(1)^2-1^3\)
\(=27000000-270000+900-1=26730899\)
So \((-299)^3=-26730899\)
3Factor the following algebraic expressions:
(i) \(4y^2+1+\frac{1}{16y^2}\)
(ii) \(9m^2-\frac{1}{25n^2}\)
(iii) \(27b^3-\frac{1}{64b^3}\)
(iv) \(x^2+\frac{5x}{6}+\frac{1}{6}\)
(v) \(27u^3-\frac{1}{125}-\frac{27u^2}{5}+\frac{9u}{25}\)
(vi) \(64y^3+\frac{1}{125}z^3\)
(vii) \(p^3+27q^3+r^3-9pqr\)
(viii) \(9m^2-12m+4\)
(ix) \(9x^3-\frac{8}{3}y^3+\frac{z^3}{3}+6xyz\)
(x) \(4x^2+9y^2+36z^2+12xy+24xz+36yz\)
(xi) \(27u^3-\frac{1}{216}-\frac{9u^2}{2}+\frac{u}{4}\)
(i) Try \(a=2y, b=\dfrac{1}{4y}\).
\(2ab=2(2y)\left(\dfrac{1}{4y}\right)=1\) and \(b^2=\dfrac{1}{16y^2}\) — both match the given terms.
So \(4y^2+1+\dfrac{1}{16y^2}=\left(2y+\dfrac{1}{4y}\right)^2\)
(ii) Difference of squares: \(9m^2-\frac{1}{25n^2}=(3m)^2-\left(\frac{1}{5n}\right)^2=\left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)\)
(iii) Difference of cubes: \(27b^3-\dfrac{1}{64b^3}=(3b)^3-\left(\dfrac{1}{4b}\right)^3\)
\(=\left(3b-\dfrac{1}{4b}\right)\left[(3b)^2+(3b)\left(\dfrac{1}{4b}\right)+\left(\dfrac{1}{4b}\right)^2\right]\)
\(=\left(3b-\dfrac{1}{4b}\right)\left(9b^2+\dfrac{3}{4}+\dfrac{1}{16b^2}\right)\)
(iv) We need \(a+b=\dfrac{5}{6}\) and \(ab=\dfrac{1}{6}\).
Trying \(a=\dfrac{1}{2}, b=\dfrac{1}{3}\): sum \(=\dfrac{5}{6}\) ✓, product \(=\dfrac{1}{6}\) ✓
So \(x^2+\dfrac{5x}{6}+\dfrac{1}{6}=\left(x+\dfrac{1}{2}\right)\left(x+\dfrac{1}{3}\right)\)
(v) Rearranged: \(27u^3-\dfrac{27u^2}{5}+\dfrac{9u}{25}-\dfrac{1}{125}\).
This matches \((a-b)^3=a^3-3a^2b+3ab^2-b^3\) with \(a=3u, b=\dfrac{1}{5}\).
\(3a^2b=3(9u^2)\left(\dfrac{1}{5}\right)=\dfrac{27u^2}{5}\)
\(3ab^2=3(3u)\left(\dfrac{1}{25}\right)=\dfrac{9u}{25}\), \(b^3=\dfrac{1}{125}\)
So \(27u^3-\dfrac{27u^2}{5}+\dfrac{9u}{25}-\dfrac{1}{125}=\left(3u-\dfrac{1}{5}\right)^3\)
(vi) Sum of cubes: \(64y^3+\dfrac{z^3}{125}=(4y)^3+\left(\dfrac{z}{5}\right)^3\)
\(=\left(4y+\dfrac{z}{5}\right)\left[(4y)^2-(4y)\left(\dfrac{z}{5}\right)+\left(\dfrac{z}{5}\right)^2\right]\)
\(=\left(4y+\dfrac{z}{5}\right)\left(16y^2-\dfrac{4yz}{5}+\dfrac{z^2}{25}\right)\)
(vii) This matches \(a^3+b^3+c^3-3abc\) with \(a=p, b=3q, c=r\).
Check: \(3abc=3(p)(3q)(r)=9pqr\), matching the given term.
So \(p^3+27q^3+r^3-9pqr=(p+3q+r)\left(p^2+9q^2+r^2-3pq-3qr-rp\right)\)
(viii) \(9m^2-12m+4=(3m)^2-2(3m)(2)+2^2=(3m-2)^2\)
(ix) Multiplying through by 3 to clear fractions:
\(3\times\left(9x^3-\dfrac{8}{3}y^3+\dfrac{z^3}{3}+6xyz\right)=27x^3-8y^3+z^3+18xyz\)
This matches \(a^3+b^3+c^3-3abc\) with \(a=3x, b=-2y, c=z\), since \(-3abc=-3(3x)(-2y)(z)=18xyz\).
\(27x^3-8y^3+z^3+18xyz=(3x-2y+z)\left(9x^2+4y^2+z^2+6xy+2yz-3xz\right)\)
Dividing back by 3: \(9x^3-\dfrac{8}{3}y^3+\dfrac{z^3}{3}+6xyz=\dfrac{1}{3}(3x-2y+z)\left(9x^2+4y^2+z^2+6xy+2yz-3xz\right)\)
(x) Try \((2x+3y+6z)^2\) with \(a=2x,b=3y,c=6z\).
\(a^2+b^2+c^2=4x^2+9y^2+36z^2\)
\(2ab=12xy\), \(2bc=36yz\), \(2ca=24xz\) — all match
So \(4x^2+9y^2+36z^2+12xy+24xz+36yz=(2x+3y+6z)^2\)
(xi) Rearranged: \(27u^3-\dfrac{9u^2}{2}+\dfrac{u}{4}-\dfrac{1}{216}\).
This matches \((a-b)^3\) with \(a=3u, b=\dfrac{1}{6}\).
\(3a^2b=3(9u^2)\left(\dfrac{1}{6}\right)=\dfrac{9u^2}{2}\)
\(3ab^2=3(3u)\left(\dfrac{1}{36}\right)=\dfrac{u}{4}\), \(b^3=\dfrac{1}{216}\)
So \(27u^3-\dfrac{9u^2}{2}+\dfrac{u}{4}-\dfrac{1}{216}=\left(3u-\dfrac{1}{6}\right)^3\)
4Simplify the following (assume denominators ≠ 0):
(i) \(\dfrac{4x^2+4x+1}{4x^2-1}\)
(ii) \(\dfrac{9(3a^3-24b^3)}{9a^2-36b^2}\)
(iii) \(\dfrac{s^3+125t^3}{s^2-2st-35t^2}\)
(i) Numerator: \(4x^2+4x+1=(2x)^2+2(2x)(1)+1^2=(2x+1)^2\). Denominator: \(4x^2-1=(2x-1)(2x+1)\).
\(\dfrac{(2x+1)^2}{(2x-1)(2x+1)}=\dfrac{2x+1}{2x-1}\)
(ii) Numerator: \(9(3a^3-24b^3)=27(a^3-8b^3)=27(a-2b)(a^2+2ab+4b^2)\)
Denominator: \(9a^2-36b^2=9(a^2-4b^2)=9(a-2b)(a+2b)\)
\(\dfrac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)}=\dfrac{3(a^2+2ab+4b^2)}{a+2b}\)
(iii) Numerator: \(s^3+125t^3=s^3+(5t)^3=(s+5t)(s^2-5st+25t^2)\)
Denominator: \(s^2-2st-35t^2\), we need \(a+b=-2, ab=-35\): \(a=-7,b=5\). So \(=(s-7t)(s+5t)\)
\(\dfrac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)}=\dfrac{s^2-5st+25t^2}{s-7t}\)
5Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units:
(i) \(25a^2-30ab+9b^2\)
(ii) \(36s^2-49t^2\)
Since area \(=\) length \(\times\) breadth, we factor each expression into two linear factors.
(i) \(25a^2-30ab+9b^2=(5a)^2-2(5a)(3b)+(3b)^2=(5a-3b)^2=(5a-3b)(5a-3b)\)
(ii) \(36s^2-49t^2=(6s)^2-(7t)^2=(6s-7t)(6s+7t)\)
6Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units:
(i) \(6a^2-24b^2\)
(ii) \(3ps^2-15ps+12p\)
Since volume \(=\) length \(\times\) breadth \(\times\) height, we factor each expression into three factors.
(i) \(6a^2-24b^2=6(a^2-4b^2)=6(a-2b)(a+2b)\)
(ii) \(3ps^2-15ps+12p=3p(s^2-5s+4)=3p(s-1)(s-4)\)
7The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
The path runs all around the outside of the square playground, so the outer boundary (playground + path) is also a square, with side \(40+2s\) metres (s metres added on each of the two opposite sides).
Area of the path = (Area of outer square) − (Area of playground) = \((40+2s)^2-40^2\)
Using \(a^2-b^2=(a+b)(a-b)\) with \(a=40+2s, b=40\):
\((40+2s)^2-40^2=\left[(40+2s)+40\right]\left[(40+2s)-40\right]=(80+2s)(2s)\)
\(=160s+4s^2\)
8If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Let the number be x. Then \(x+\dfrac{1}{x}=\dfrac{10}{3}\)
Multiplying both sides by \(3x\): \(3x^2+3=10x\)
\(3x^2-10x+3=0\)
Splitting the middle term (we need two numbers multiplying to \(3\times3=9\) and adding to \(-10\): these are \(-9\) and \(-1\)):
\(3x^2-9x-x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
So \(x=\dfrac{1}{3}\) or \(x=3\)
Check: \(3+\dfrac{1}{3}=\dfrac{10}{3}\) ✓ and \(\dfrac{1}{3}+3=\dfrac{10}{3}\) ✓ — both give the same pair of reciprocal numbers.
9A rectangular pool has area \(2x^2+7x+3\) square hastas. If its width is \(2x+1\) hastas, find its length. (Hasta was a unit used to measure length.)
Since area \(=\) length \(\times\) width, we factor \(2x^2+7x+3\) and identify the factor matching the given width.
We need two numbers multiplying to \(2\times3=6\) and adding to 7: these are 6 and 1.
\(2x^2+7x+3=2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)\)
So area \(=(2x+1)(x+3)\). Since width \(=2x+1\), the length must be the other factor.
10*If both \(x-2\) and \(x-\frac{1}{2}\) are factors of \(px^2+5x+r\), show that \(p=r\).
If \((x-2)\) is a factor, then substituting \(x=2\) into \(px^2+5x+r\) must give 0:
\(p(2)^2+5(2)+r=0\)
\(4p+10+r=0\)
\(4p+r=-10\) …(1)
If \(\left(x-\frac{1}{2}\right)\) is a factor, then substituting \(x=\frac{1}{2}\) must also give 0:
\(p\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+r=0\)
\(\frac{p}{4}+\frac{5}{2}+r=0\)
Multiplying by 4: \(p+10+4r=0\)
\(p+4r=-10\) …(2)
Subtracting (2) from (1): \((4p+r)-(p+4r)=-10-(-10)\)
\(3p-3r=0\)
\(3(p-r)=0\)
\(p=r\)
11*If \(a+b+c=5\) and \(ab+bc+ca=10\), then prove that \(a^3+b^3+c^3-3abc=-25\).
\((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\)
\(5^2=a^2+b^2+c^2+2(10)\)
\(a^2+b^2+c^2=25-20=5\)
\(a^2+b^2+c^2-ab-bc-ca=5-10=-5\)
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(5)(-5)=-25\)
12*By factoring the expression, check that \(n^3-n\) is always divisible by 6 for all natural numbers n. Give reasons.
\(n^3-n=n(n^2-1)=n(n-1)(n+1)=(n-1)\,n\,(n+1)\)
This is a product of 3 consecutive integers. Among any 3 consecutive integers, at least one is a multiple of 3, and at least one is even.
\(\therefore (n-1)n(n+1)\) is divisible by both 2 and 3, hence by \(2\times3=6\).
13*Find the value of:
(i) \(x^3+y^3-12xy+64\), when \(x+y=-4\)
(ii) \(x^3-8y^3-36xy-216\), when \(x=2y+6\)
(i) With \(a=x, b=y, c=4\): \(-3abc=-3(x)(y)(4)=-12xy\), matching the given term.
\(x^3+y^3+4^3-3(x)(y)(4)=(x+y+4)(x^2+y^2+16-xy-4y-4x)\)
Given \(x+y=-4\): \(x+y+4=0\)
\(\therefore x^3+y^3-12xy+64=0\)
(ii) With \(a=x, b=-2y, c=-6\): \(-3abc=-3(x)(-2y)(-6)=-36xy\), matching the given term.
\(x^3+(-2y)^3+(-6)^3-3(x)(-2y)(-6)=(x-2y-6)(\text{quadratic factor})\)
Given \(x=2y+6\): \(x-2y-6=0\)
\(\therefore x^3-8y^3-36xy-216=0\)
Frequently Asked Questions
For example, \(x^2-1=24\) is true only for \(x=5\) or \(x=-5\), so it is an equation, while \((x+y)^2=x^2+2xy+y^2\) is true for every x and y, so it is an identity.
\((x+y)^2=x^2+2xy+y^2\)
\((x-y)^2=x^2-2xy+y^2\)
\((x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\)
\((x+y)(x-y)=x^2-y^2\)
\((x+a)(x+b)=x^2+(a+b)x+ab\)
the cube identities \((x+y)^3\) and \((x-y)^3\)
the sum/difference of cubes \(x^3\pm y^3\)
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\)
Want quick revision before your test?
Class 9 Ganita Manjari Formulas — every formula and identity from all 8 chapters of Part I, all captured on just one printable sheet. Get it now to make your revision easy and comprehensive.