## Monday, 6 June 2011

### Algebraic Expressions and Identities soved

1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy
Answer: Term: xyz2 -- Coefficient = 5
Term: zy –- Coefficient = 3
(ii) 1 + x + x2
Answer: Term: x – Coefficient = 1 & Term: x2 – Coefficient = 1
(iii) 4x2y2 – 4x2y2z2 + z2
Answer: 4 is the coefficient for x2y2z2
1 is the term for z2
(iv) 3 – pq + qr – rp
Answer: For each term the coefficient is 1
(v) 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?
x + y: Binomial
1000: Mononomial
x + x2 + x3 + x4: Polynomial
7 + y + 5x: Binomial
2y – 3y2: Binomial
2y – 3y2 + 4y3: Trinomial
5x – 4y + 3xy: Trinomial
4z – 15z2: Binomial
ab + bc + cd + da: Polynomial
pqr: Mononomial
p2q + pq2: Binomial
2p + 2q: Binomial
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Answer: (ab - bc) + (bc - ca) + (ca-ab)
ab + bc + ca - bc - ca - ab =
= 0
(ii) a – b + ab, b – c + bc, c – a + ac
Answer: (a - b + ab) + (b - c + bc) + (c - a + ac)
= a + b + c + ab + bc + ca - b - c - a
= ab + bc + ca
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
= (2p2q2 - 3pq + 4) + (5 + 7pq - 3p2q2)
= 2p2q2 - 3p2q2 - 3pq + 7pq + 4 + 5
= - p2q2 + 4pq + 9
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Answer: (l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 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)
= 5xy - 2yz - 2zx + 10xyz - 3xy - 5yz + 7zx
= 2xy - 7yz + 5zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10
from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Answer: (18 - 3p - 11q + 5pq - 2pq2 + 5p2q) - (4p2q - 3pq + 5pq2 - 8p + 7q - 10)
= 18 - 3p - 11q + 5pq - 2pq2 + 5p2q - 4p2q + 3pq - 5pq2 + 8p - 7q + 10
= 28 + 5p - 18q + 8pq - 7pq2 - p2

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)
Answer: (p2 - pq) + (q2 - qr) + (r2 - pr)
= p2 + q2 + r2 - pq - qr - pr
(b) Add: 2x (z – x – y) and 2y (z – y – x)
Answer: (2xz - 2x2 - 2xy) + (2yz - 2y2 - 2xy)
= 2xz - 4xy + 2yz - 2x2 - 2y2
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
Answer: (40ln - 12lm + 8l2) - (3l2 - 12lm + 15ln)
= 40ln - 12lm + 8l2 - 3l2 - 12lm + 15ln
= 55ln - 24lm + 5l2
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
= (-4ac + 4bc + 4c2) - (3a2 + 3ab + 3ac)
= -4ac + 4bc + 4c2 - 3a2 - 3ab - 3ac
= -7ac + 4bc + 4c2 - 3a2 - 3ab
6. Multiply the binomials.
i) (2x + 5) and (4x – 3)
Answer: (2x + 5)(4x - 3)
= 2x x 4x - 2x x 3 + 5 x 4x - 5 x 3
= 8x² - 6x + 20x -15
= 8x² + 14x -15
(ii) (y – 8) and (3y – 4)
Answer: ( y - 8)(3y - 4)
= y x 3y - 4y - 8 x 3y + 32
= 3y2 - 4y - 24y + 32
= 3y2 - 28y + 32
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
Answer: (2.5l - 0.5m)(2.5l + 0.5)
Using (a+b)(a-b) = a2 - b2
We get = 6.25l2 - 0.25m2
(iv) (a + 3b) and (x + 5)
Answer: = ax + 5a + 3bx + 15b
(v) (2pq + 3q2) and (3pq – 2q2)
Answer: = 2pq x 3pq - 2pq x 2q2 + 3q2 x 3pq - 3q2 x 2q2
= 6p2q2 - 4pq3 + 9pq3 - 6q4
= 6p2q2 - 5pq3 - 6p4

7. Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)

iii) (a2+ b) (a + b2
(iv) (p2– q2) (2p + q)

Answer: = 15 + 5x - 6x - 2x2
= 15 - x - x 2
(ii) (x + 7y) (7x – y)
Answer: = 7x2 - xy + 49xy - 7y2
= 7x2 - 7y2 + 48xy
iii) (a2+ b) (a + b2)
Answer: a2 x a + a2 x b + a x b + b3
= a3 + a2b + ab + b3
= a3 + b3 + a2b + ab
(iv) (p2– q2) (2p + q)
Answer: = 2p3 + p2q - 2pq2 - q3
= 2p3 - q3 + p2q - 2pq2
8. Simplify.
(i) (x2– 5) (x + 5) + 25
Answer: = x3 + 5x2 - 5x - 25 + 25
= x3 + 5x2 -5x

(ii) (a2+ 5) (b3+ 3) + 5
Answer: a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 5b3 + 3a2 + 20

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
Answer: = (ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)
= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd
= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)
Answer: 2x2 + xy + 2xy + y2 + x2 - xy + 2xy - 2y2
= 3x2 + 4xy - y2

(vi) (x + y)(x2– xy + y2)
Answer: = x3 - x2y + xy2 + x2y - xy2 + y3
= x3 + y3

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
Answer: = 2.25x2 + 6xy + 4.5x - 6xy - 16y2 - 12y - 4.5x + 12y
= 2.25x2 - 16y2

(viii) (a + b + c)(a + b – c)
Answer: = a2 + ab - ac + ab + b2 - bc + ac + bc - c2
= a2 + b2 - c2 + 2ab

9 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
(iii) (2a – 7) (2a – 7)
Answer: Using (a - b)2 = a2 - 2ab + b2 we get the following equation:
= 4a2 - 28a + 49

10. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
Answer: x2 + (3+7)x + 21
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1)
= 16x2 + (-5-1)4x + 5
= 16x2 - 20x + 5

(iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x - 5
= 16x2 +16x - 5

(v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)4x + 15y2
= 4x2 + 32xy + 15y2

(vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz - 8
= x2y2z2 - 6xyz - 8
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#### 1 comment:

1. Variable and expression are must be understood while solving it's related problems.A symbol for a number which is usually a letter like x or y called as Variable and a number on its own is called a Constant.When these two or more variables are joined together with the help of the symbols are called expressions.
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