Algebraic Expressions 1. Expressions are formed from variables and constants. 2. Terms are added to form expressions. Terms themselves are formed as product of factors. 3. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials respectively. In general, any expression containing one or more terms with non-zero coefficients (and with variables having non- negative exponents) is called a polynomial. 4. Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same. 5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. 6. There are number of situations in which we need to multiply algebraic expressions: for example, in finding area of a rectangle, the sides of which are given as expressions. 7. A monomial multiplied by a monomial always gives a monomial. 8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial. 9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial). Note that in such multiplication, we may get terms in the product which are like and have to be combined. 10. An identity is an equality, which is true for all values of the variables in the equality. On the other hand, an equation is true only for certain values of its variables. An equation is not an identity. 11. The following are the standard identities: (a + b) 2 = a 2 + 2ab + b 2..(I) (a b) 2 = a 2 2ab + b 2.(II) (a + b) (a b) = a 2 b 2.(III) 12. Another useful identity is (x + a) (x + b) = x 2 + (a + b) x + ab (IV) 13. The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.
Exercise 1 1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz 2 3zy Answer: Term: xyz 2 -- Coefficient = 5 Term: zy Coefficient = 3 (ii) 1 + x + x 2 Answer: Term: x Coefficient = 1 & Term: x 2 Coefficient = 1 (iii) 4x 2 y 2 4x 2 y 2 z 2 + z 2 Answer: 4 is the coefficient for x 2 y 2 z 2 1 is the term for z 2 (iv) 3 pq + qr rp Answer: For each term the coefficient is 1 x y (v) + xy 2 2 Answer: 2 1 is the coefficient for x and y and for xy it is 1 (vi) 0.3a 0.6ab + 0.5b Answer: 0.3 is the coefficient for a, for b there are two coefficients 0.6 and 0.5 2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? Answer: x + y: Binomial 1000: Mononomial x + x 2 + x 3 + x 4 : Polynomial 7 + y + 5x: Binomial 2y 3y 2 : Binomial 2y 3y 2 + 4y 3 : Trinomial 5x 4y + 3xy: Trinomial 4z 15z 2 : Binomial ab + bc + cd + da: Polynomial pqr: Mononomial p 2 q + pq 2 : Binomial 2p + 2q: Binomial
3. Add the following. (i) ab bc, bc ca, ca ab Answer: ( ab bc) + ( bc ca) + ( ca ab) = ab bc + bc ca + ca ab = ab ab + bc + bc ca + ca = bc + bc = 2bc (ii) a b + ab, b c + bc, c a + ac Answer: ( a b + ab) + ( b c + bc) + ( c a + ac) = a b + ab + b c + bc + c a + ac = a a b + b c + c + ab + bc + ac = ab + bc + ca (iii) 2p 2 q 2 3pq + 4, 5 + 7pq 3p 2 q 2 Answer: ( 2 p²q² - 3pq + 4) + (5 + 7pq - 3p²q²) = 2 p²q² - 3pq + 4 + 5 + 7pq - 3p²q² = 2 p²q² - 3p²q² - 3pq + 7pq + 4 + 5 = p²q² + 4pq + 9 = 4 pq + 9 p²q² (iv) l 2 + m 2, m 2 + n 2, n 2 + l 2, 2lm + 2mn + 2nl Answer: l² + m² + m² + n² + n² + l² + 2lm + 2mn + = 2 l² + 2m² + 2n² + 2lm + 2mn + 2nl = 2 ( l² + m² + n² + lm + mn + nl) 2nl 4. (a) Subtract 4a 7ab + 3b + 12 from 12a 9ab + 5b 3 Answer: ( 12a 9ab + 5b 3) (4a 7ab + 3b + 12) = 12a 9ab + 5b 3 4a + 7ab 3b 12 (signs are reversed after sign once bracket is opened) = 8a 2ab + 2b 15 (b) Subtract 3xy + 5yz 7zx from 5xy 2yz 2zx + 10xyz Answer: ( 5xy 2yz 2zx + 10xyz) (3xy + 5yz 7zx) = 5 xy 2yz 2zx + 10xyz 3xy 5yz + 7zx = 2 xy 7yz + 5zx + 10xyz
(c) Subtract 4p 2 q 3pq + 5pq 2 8p + 7q 10 from 18 3p 11q + 5pq 2pq 2 + 5p 2 q Answer: ( 18 3p 11q + 5pq 2 pq² + 5p²q) - (4p²q - 3pq + 5pq² - 8p + 7q = 18 + 3p 11q + 5pq 2 pq² + 5p²q - 4p²q + 3pq - 5pq² + 8p - 7q + 10 = 28 + 11p 18q + 8pq7 pq² + p²q 10)
Exercise 2 1. Find the product of the following pairs of monomials. (i) 4, 7p Answer: 4 7 p = 28p (ii) 4p, 7p Answer: 4 p 7 p = 28p² (iii) 4p, 7pq Answer: 4 p 7 pq = 28p²q (iv) 4p3, 3p Answer: 4 p³ 3p = 12 p (v) 4p, 0 Answer: 4 p 0 = 0 2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively. (p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np) Answer: Area = Length breadth (i) p q = pq (ii) 10 m 5n = 50mn (iii) 20 x² 5y² = 100x²y² (iv) 4 x 3x² = 12x ³ (v) 3mn 4np = 12mn²p 3. Complete the following table of products: First Mononomial 2x -5y 3x² -4xy 7x²y -9x²y² Second Mononomial 2x 4x² -10xy 6x³ -8x²y 14x³y -18x³y² -5y -10xy 10y² -15x²y 20xy² -35x²y² 45x²y³ 3x² 6x³ -15x²y 9x -12x³y 21x y² -27x y² -4xy -8x²y 20xy² -12x³y 16x²y² -28x³y² 36x³y³ 7x²y 14x³y -35x²y² 21x y² -28x³y² 49x y² -63x y³ -9x²y² -18x³y² 45x²y³ -27x y² 36x³y³ -63x y³ 81x y
4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively. (i) 5a, 3a², 7a (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy² (iv) a, 2b, 3c Answer: Volume= length breadth height (i) 5a 3a² 7a =105a 7 (ii) 2p 4q 8r = 64pqr (iii) xy 2x²y 2xy² = 4x y (iv) a 2b 3c = 6abc 5. Obtain the product of (i) xy, yz, zx (ii) a, a², a³ (iii) 2, 4y, 8y², 16y³ (iv) a, 2b, 3c, 6abc (v) m, mn, mnp Answer: (i) x²y²z² (ii) a 5 (iii) 1024y 6 (iv) 36a²b²c² (v) -m³n²p
Exercise 3 1. Carry out the multiplication of the expressions in each of the following pairs. (i) 4p, q + r Answer: 4 p( q + r) = 4 pq + 4 pr (ii) ab, a b Answer: ab ( a b) = a²b - ab² (iii) a + b, 7a²b² Answer: ( a + b)7a²b² = 7a³b²+7a²b³ (iv) a² 9, 4a Answer: (a² - 9)4a = 4a³-36a (v) pq + qr + rp, 0 Answer: ( pq + qr + rp)0 = 0 2. Find the product. (i) (a²) (2a 22 ) (4a 26 ) Answer: As you know a n a m a o = a m+n+o So, (a²) (2a 22 ) (4a 26 ) = 8a 50 2 9 (ii) xy x ²y² 3 10 3 = x ³y³ 5 10 6 (iii) pq ³ p³q 3 5 =-4p q (iv) x x² x³ x =x 14
3. (a) Simplify 3x (4x 5) + 3 and find its values for (i) x = 3 (ii) x = 2 1 Answer: 3x(4x-5)+3 =12x²-15x+3 (i) putting x=3 in the equation we get 12 3²-15 3+3 =108-45+3 =63 (ii) putting x= 2 1 in the equation we get 12 4 1-15 2 1 +3 = 6 15 =3- +3 2 15 + 2 6 = 15 2 = 1 7 2 (b) Simplify a (a² + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = 1. Answer: a(a²+a+1) =a³+a²+a (i) putting a= 0 in the equation we get 0³+0²+0=0 (ii) putting a=1 in the equation we get 1³+1²+1=1+1+1=3 (iii) putting a=-1 in the equation we get -1³+1²+1=-1+1+1=1 5. (a) Add: p ( p q), q ( q r) and r ( r p) Answer: p( p q) + q( q r) + r( r p) = p² - pq + q² - qr + r² - rp = p² + q² + r² - pq - qr - pr (b) Add: 2x (z x y) and 2y (z y x) Answer: 2x( z x y) + 2y( z y x) = 2 xz 2x² - 2xy + 2yz - 2y² - 2xy = 2xz 2xy + 2yz 2x² - 2y²
(c) Subtract: 3l (l 4 m + 5 n) from 4l ( 10 n 3 m + 2 l ) Answer: 4l (10n 3m + 2l) 3l( l 4m + 5n) = 40 ln 12lm + 8l² - 3l² + 12lm -15ln = 25lm + 5l² (d) Subtract: 3a (a + b + c ) 2 b (a b + c) from 4c ( a + b + c ) Answer: 4c ( a + b + c) ( 3a( a + b + c) 2b( a b + c) ) = 4 ac + 4bc + 4c² - (3a² + 3ab + 3ac - 2ab + 2b² - 2bc) = 4 ac + 4bc + 4c² - 3a² - 3ab - 3ac + 2ab - 2b² + 2bc = 7ac + 6bc + 4c² - 3a² - ab - b² = 4 c² - 3a² - 2b² - ab + 6bc - 7ac
= 6 2. 5 ² l - 0.2 5 m² a 3b ) ( 5 Finish Line & Beyond Exercise 4 1. Multiply the binomials. (i) (2x + 5) and (4x 3) Answer: ( 2x + 5)(4x 3) = 2 x 4x 2x 3 + 5 4x 5 = 8 x² - 6x + 20x -15 = 8 x² + 14x -15 3 (ii) (y 8) and (3y 4) Answer: ( y 8)(3y 4) = y 3 y y 4 8 3y + 8 = 3 y² - 4y - 24y + 32 = 3 y² - 28y + 32 4 (iii) (2.5l 0.5m) and (2.5l + 0.5m) Answer: ( 2.5l 0.5m)(2.5l + 0.5m) This can be solved using following identity: ( a b)( a + b) = a² - b² ; Assume 2.5l = a and 0.5m = b Resulting Equation = 2.5²l² - 0.5²m² (iv) (a + 3b) and (x + 5) Answer: ( + x + ) = a x + a 5 + 3b x + 3b = ax + 5 a + 3bx + 15b 5 (v) (2pq + 3q²) and (3pq 2q²) Answer: ( 2 pq + 3q²)(3pq - 2q²) = 2 pq 3pq 2 pq 2q² + 3q² 3pq - 3q² = 6 p²q² - 4pq³ + 9pq³ - 9q =6p²q²+5pq³-9q 3q² 3 2 (vi) a ² + 3b² 4 a² - b² 4 3 3 8 = a² + 3b² 4a² + b² 4 3
3 3 8 = a ² 4a² - a² b² + 3b² 4a² - 3b² 4 4 3 = 3a -2a²b²+12a²b²-8b =3a +10a²b²-8b 8 b² 3 2. Find the product. (i) (5 2x) (3 + x) Answer: ( 5 2x )(3 + x) = 5 3 + 5 x 2x 3 2x = 15 + 5x 6x 2x² = 15 x 2x² x (ii) (x + 7y) (7x y) Answer: ( x + 7y)(7x y) = x 7x x y + 7y 7x 7y = 7 x² - xy + 49xy - 7y² = 7 x² + 48xy - 7y² y (iii) (a² + b) (a + b²) Answer: ( a² + b)(a + b²) = a² a + a² b² + b a + b b² = a³ + a²b² + ab + b³ = a³ + b³ + a²b² + ab (iv) (p² q²) (2p + q) Answer: ( p² - q²)(2p + q) = p² 2p + p²q - q² 2p - q² = 2 p ³ + p²q - 2pq² - q³ q 3. Simplify. (i) (x² 5) (x + 5) + 25 Answer: ( x² - 5)(x + 5) + 25 = ( x² x + 5 x - 5 x - 5 5) + = x³ + 5x 5x 25 + 25 = x³ 25 (ii) (a² + 5) (b³ + 3) + 5 Answer: ( a² + 5)( b³ + 3) + 5 = a² b³ + a² 3 + 5 b³ + 5 3 = a² b³ + 3a² + 5b³ + 15
(iii) (t + s²) (t² s) Answer: ( t + s²)(t² - s) = t t² - t s + s² t² - s² = t ³ ts + s²t² - s³ s (iv) (a + b) (c d) + (a b) (c + d) + 2 (ac + bd) Answer: ( a + b)( c d) + ( a b)( c + d) + 2( ac + bd) = ( ac ad + bc bd) + ( ac + ad bc bd) + 2ac + 2bd = ac ad + bc bd + ac + ad bc bd + 2 ac + 2bd = 4ac (v) (x + y)(2x + y) + (x + 2y)(x y) Answer: ( x + y)(2x + y) + ( x + 2y)( x y) = ( 2x² + xy + 2xy + y²) + (x² - xy + 2xy - 2y²) = 2 x² + 3xy + y² + x² + xy - 2y² = 3 x² + 4xy - y² (vi) (x + y)(x² xy + y²) Answer: ( x + y)( x² - xy + y²) = x³ x²y + xy² + x²y - xy² + y³ = x³ + y³ (vii) (1.5x 4y)(1.5x + 4y + 3) 4.5x + 12y Answer: ( 1.5x 4y)(1.5 x + 4y + 3) 4.5x + 12y = ( 2.25x² + 6xy + 4.5x - 6xy -16y² -12y) - 4.5x + 12y = 2.25x² + 4.5x -16y² -12y - 4.5x + 12y = 2.25x² -16y² (viii) (a + b + c)(a + b c) Answer: ( a + b + c)( a + b c) = a² + ab - ac + ab + b² - bc + ac + = a² + b² - c² + 2ab bc - c²
2a )7 ² = 4a ² - 28 a Finish Line & Beyond Exercise 5 1. Use a suitable identity to get each of the following products. (i) (x + 3) (x + 3) Answer: Using (a+b) ²=a²+2ab+b², we get the following equation: x² + 6x + 9 (ii) (2y + 5) (2y + 5) Answer: 4y² + 20y + 25 (same as previous question) (iii) (2a 7) (2a 7) Answer: As you know ( a b)² = a² - 2ab + b² So, ( + 49 1 (iv) 3a 3a 2 1 2 1 Answer: 9 a ² - 3a + (same as previous question) 4 (v) (1.1m 0.4) (1.1m + 0.4) Answer: Use ( a b)( a + b) = a² - b² 1.21m²-0.16 (vi) (a² + b²) ( a² + b²) Answer: (a²+b²)(-a²+b²) =(b²+a²)(b²-a²) =b -a (same as previous question) (vii) (6x 7) (6x + 7) Answer: 36x²-49 (same as previous question) (viii) ( a + c) ( a + c) Answer: (-a+c)(-a+c) =(c-a)(c-a) =c²+a²-2ac
(ix) x + 2 3y 4 x + 2 3y 4 Answer: Use (a+b) ² x² 9y² 3 = + + xy 4 16 4 (x) (7a 9b) (7a 9b) Answer: Using (a-b) ² = 49 a² + 81b² -126ab 2. Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products. (i) (x + 3) (x + 7) Answer: ( x + 3)( x + 7) = x² + (3 + 7)x + 21 = x² + 10x + 21 (ii) (4x + 5) (4x + 1) Answer: ( 4x + 5)(4x + 1) = 16 x² + (5 + 1)4x + 5 = 16 x² + 24x + 5 (iii) (4x 5) (4x 1) Answer: ( 4x 5)(4x 1) = 16 x² + (-5 + -1)4x + (-5) = 16x² - 24x - 5 (iv) (4x + 5) (4x 1) Answer: ( 4x + 5)(4x 1) = 16 x² + (5-1)4x - 5 = 16 x² + 4x - 5 (v) (2x + 5y) (2x + 3y) Answer: ( 2x + 5y)(2x + 3y) = 4 x² + (5y + 3y)2x + 15y² = 4 x² + 8xy + 15y²
(vi) (2a² + 9) (2a² + 5) Answer: ( 2a² + 9)(2a² + 5) =2a +(9+5)2a²+45 =2a +28a²+45 (vii) (xyz 4) (xyz 2) Answer: ( xyz 4)( xyz 2) = x²y²z² + (-4-2)xyz + 8 = x²y²z² - 6xyz + 8 3. Find the following squares by using the identities. (i) (b 7) ² Answer: Use (a-b) ²=a²+b²-2ab ( b 7)² = b² + 49-14b (ii) (xy + 3z) ² Answer: Use ( a + b)² = a² + b² + ( xy + 3z)² = x²y² + 9z² + 6xyz 2ab (iii) (6x² 5y) ² Answer: Use ( a b)² = a² + b² - 2ab ( 6x ² - 5y)² =36x +25y²-60x²y 2 3 (iv) ( m + n) ² 3 2 Answer: Use ( a + b)² = a² + b² + 2ab 2 3 4 9 2 3 ( m + n)² = m² + n² + 2 3 2 9 4 3 2 4 9 = m² + n² + 2mn 9 4 mn (v) (0.4p 0.5q) ² Answer: Use ( a b)² = a² + b² + 2ab ( 0.4 p 0.5q)² = 0.16p² + 0.25q² - 0.4pq
(vi) (2xy + 5y) ² Answer: Use ( a + b)² = a² + b² + 2ab ( 2xy + 5y)² = 4x²y² + 25y² + 20xy² 4. Simplify. (i) (a² b²)² Answer: Use ( a b)² = a² + b² + (a²-b²)²=a +b -2a²b² 2ab (ii) (2x + 5) ² (2x 5) ² Answer: ( 2x + 5)² - (2x - 5)² = ( 4x² + 25 + 20x) - (4x² + 25-20x) = 4 x² + 25 + 20x - 4x² - 25 + 20x = 20x (iii) (7m 8n) ² + (7m + 8n) ² Answer: ( 7m 8n)² + (7m + 8n)² = ( 49m² + 64n² -112mn) + (49m² + 64n² -112mn) = 49 m² + 64n² + 112mn + 49m² + 64n² -112mn = 98 m² + 128n² (iv) (4m + 5n) ² + (5m + 4n) ² Answer: ( 4m + 5n)² + (5m + 4n)² = 16 m² + 25n² + 40mn + 25m² + 16n² + = 41 m² + 41n² + 8 40mn (v) (2.5p 1.5q) ² (1.5p 2.5q) ² Answer: ( 2.5 p 1.5q)² - (1.5p - 2.5q)² = ( 6.25 p² + 2.25q² - 0.75pq) - (2.25p² + 6.25q² - 0.75pq) = 6.25p² + 2.25q² - 0.75pq - 2.25p² - 6.25q² + 0.75pq = 4 p² - 4q² (vi) (ab + bc) ² 2ab²c Answer: ( ab + bc)² - 2ab²c = ( a²b² + b²c² + 2ab²c) - 2ab²c = a²b² + b²c²
(vii) (m² n²m) ² + 2m³n² Answer: ( m² - n²m)² + 2m³n² =m +n m²-2m n²+2m³n² =m +n m² 5. Show that. (i) (3x + 7)² 84x = (3x 7) ² Answer: LHS: ( 3x + 7)² - 84x = 9 x² + 49 + 42x - 84x = 9 x ² + 49-42x..(1) RHS ( 3x 7) ² = 9 x ² + 49-42x (2) Equations (1) & (2) prove that LHS =RHS (ii) (9p 5q) ² + 180pq = (9p + 5q) ² Answer: LHS: ( 9 p 5q)² + 180pq = 81 p² + 25q² - 90pq + 180pq = 81 p ² + 25q² + 90pq.(1) RHS: ( 9 p + 5q) ² = 81 p ² + 25q² + 90pq..(2) Equations (1) & (2) prove that LHS = RHS 4 3 16 9 (iii) ( m n)² + 2mn = m² + n² 3 4 9 16 4 3 Answer: LHS: ( m n)² + 2mn 3 4 16 9 = m² + n² - 2mn + 2mn 9 16 16 9 = m ² + n² = RHS Proved 9 16 (iv) (4pq + 3q) ² (4pq 3q) ² = 48pq² Answer: LHS: ( 4 pq + 3q)² - (4pq - 3q)² = ( 16 p²q² + 9q² + 24pq²) - (16p²q² + 9q² - 24pq²) = 16 p²q² + 9q² + 24pq² -16p²q² - 9q² + 24pq² = 48pq² = RHS Proved
(v) (a b) (a + b) + (b c) (b + c) + (c a) (c + a) = 0 Answer: ( a b)( a + b) + ( b c)( b + c) + ( c a)( c + a) = a ² - b² + b² - c² + c² - a² = 0 Proved 6. Using identities, evaluate. (i) 71² Answer: 71 ² Can be written as follows: ( 70 + 1)² which can be solved using ( a + b)² = a² + b² + 2ab = 70 ² + 1² + 2 70 1 = 4900 + 1 + 140 = 5041 (ii) 99² Answer: 99² can be written as follows: (100-1) ² and can be solved using ( a b)² = a² + b² - 2ab = 100 ² + 1² - 2 100 1 = 10000 + 1 200 = 99801 (iii) 102² Answer: 102 ² = (100 + 2)² = 100 ² + 2² + 2 100 2 = 10000 + 4 + 400 = 10404 (iv) 998² Answer: 998 ² = (1000-2)² = 1000 ² + 2² - 2 1000 2 = 1000000 + 4 4000 = 996004 (v) 5.2² Answer: 5.2² = (5 + 0.2)² = 5 ² + 0.2² + 2 5 0.2 = 25 + 0.04 + 2 = 27.04 (vi) 297 303 Answer: ( 297 303) = (300 3)(300 + 3) Using ( a b)( a + b) = a² - b² we get
300² - 3² = 90000 9 = 89991 (vii) 78 82 Answer: 78 82 = (80 2)(80 + 2) = 80² - 2² = 6400 4 = 6396 (viii) 8.9² Answer: 8.9² = (9-0.1)² = 9 ² + 0.1² - 2 9 0.1 = 81 + 0.01 1.8 = 79.21 (ix) 1.05 9.5 Answer: 10.5 9.5 = (10 + 0.5)(10 0.5) = 10² - 0.5² = 100 0. 25 = 99. 75 7. Using a² b² = (a + b) (a b), find (i) 51² 49² Answer: 51² - 49² = ( 51 + 49)(51 49) = 100 2 = 200 (ii) (1.02) ² (0.98) ² Answer: (1.02)² - (0.98)² = ( 1.02 + 0.98)(1.02 0.98) = 2 0.04 = 0.08 (iii) 153² 147² Answer: 153² -147² = ( 153 + 147)(153 147) = 300 6 = 1800 (iv) 12.1² 7.9² Answer: 12.1² - 7.9² = ( 12.1 + 7.9)(12.1 7.9) = 20 4.2 = 84
8. Using (x + a) (x + b) = x² + (a + b) x + ab, find (i) 103 104 Answer: 103 104 = ( 100 + 3)(100 + 4) = 100 ² + (3 + 4)100 + 4 = 10000 + 700 + 12 = 10712 3 (ii) 5.1 5.2 Answer: 5.1 5. 2 = ( 5 + 0.1)(5 + 0.2) = 5 ² + (0.1 + 0.2)5 + 0.1 = 25 + 1.5 + 0.02 = 26.52 0.2 (iii) 103 98 Answer: 103 98 = ( 90 + 13)(90 + 8) = 90 ² + (13 + 8)90 + 13 = 8100 + 1890 + 104 = 10094 8 (iv) 9.7 9.8 Answer: 9.7 9. 8 = ( 9 + 0.7)(9 + 0.8) = 9 ² + (0.7 + 0.8)9 + 0.7 = 81 + 13.5 + 0.56 = 95.05 0.8