Exercise 9.2
Question 1. Find the product of the following pairs of monomials.
(i) 4, 7p (ii) â€“4p, 7p (iii) â€“4p, 7pq
(iv) 4p^{3}, â€“3p (v) 4p, 0
Solution:
(i) 4 and 7p
4 * 7p = (4 * 7) * p = 28p
(ii) â€“4p and 7p
â€“4p * 7p = {(â€“4) * 7} * p * p = â€“28p^{2}
(iii) â€“4p and 7pq
â€“4p * 7pq = (â€“4 * 7) * p * qp = â€“28 * p^{2}q = â€“28p^{2}q
(iv) 4p^{3} and â€“3p
4p^{3} * (â€“3p) = {4 * (â€“3)}p^{3} * p = â€“12 * p^{4} = â€“12p^{4}
(v) 4p and 0 â‡’ 4p * 0 = 0
Question 2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x^{2}, 5y^{2}); (4*, 3x^{2}); (3mn, 4np)
Solution:
(i) Length = p Breadth = q} âˆ´ Area of the rectangle = p * q = pq
(ii) Length = 10m Breadth = 5n} âˆ´ Area = 10m * 5n = 10 * 5 * m * n = 50mn
(iii) Length = 20x^{2} Breadth = 5y^{2}} âˆ´ Area = 20x^{2} * 5y^{2 }= 20 * 5 * x^{2} * y^{2} = 100x^{2}y^{2}
(iv) Length = 4x Breadth = 3*^{2}} âˆ´ Area = 4x * 3 * x^{2 }= 4 * 3 * x *x^{2}= 12*^{3}
(v) Length = 3mn Breadth = 4np} âˆ´ Area = 3mn * 4np = (3 * 4) * m * n * n * p = 12 mn^{2}p
Question 3. Complete the table of products.
First monomial â†’ Second monomial | 2* | â€“5y | 3*^{2} | â€“4*y | 7*^{2}y | â€“9x^{2}y^{2} |
2x â€“5y 3x^{3} â€“4xy 7x^{2}y â€“9x^{2}y^{2} | 4*^{2} â€¦ â€¦ â€¦ â€¦ â€¦ | â€¦ â€¦ â€¦ â€¦ â€¦ â€¦ | â€¦ â€“15x^{2}y â€¦ â€¦ â€¦ â€¦ | â€¦ â€¦ â€¦ â€¦ â€¦ â€¦ | â€¦ â€¦ â€¦ â€¦ â€¦ â€¦ | â€¦ â€¦ â€¦ â€¦ â€¦ â€¦ |
Solution: We have
First monomial â†’ Second monomial | 2x | -5y | 3x^{2} | -4xy | 7x^{2}y | â€“9x^{2}y^{2} |
2x | 4x^{2} | -10xy | 6x^{3} | -8x^{2}y | 14x^{3}y | -18x^{3}y^{2} |
-5y | -10xy | 25y^{2} | -15x^{2}y | 20xy^{2} | -35x^{2}y^{2} | 45x^{2}y^{3} |
3x^{3} | 6x^{3} | -15x^{2}y | 9x^{4} | -12x^{3}y | 21x^{4}y | -27x^{4}y^{2} |
-4xy | -8x^{2}y | 20xy^{2} | -12x^{3}y | 16x^{2}y^{2} | -28x^{3}y^{2} | 36x^{3}y^{3} |
7x^{2}y | 14x^{3}y | -35x^{2}y^{2} | 21x^{4}y | -28x^{3}y^{2} | 49x^{4}y^{2} | -63x^{4}y^{3} |
-9x^{2}y^{2} | -18x^{3}y^{2} | 45x^{2}y^{3} | -27x^{4}y^{2} | 36x^{3}y^{3} | -63x^{4}y^{3} | 81x^{4}y^{4} |
Question 4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a^{2}, 7a^{4} (ii) 2p, 4q, 8r (iii) xy, 2x^{2}y, 2xy^{2} (iv) a, 2b, 3c
Solution: Volume of a rectangular box = Length * Breadth * Height
(i) Length = 5a; Breadth = 3a^{2}, Height = 7a^{4}
âˆ´ Volume = Length Ã— Breadth Ã— Height
= 5a * 3a^{2} * 7a^{4}
= (5 * 3 * 7) * a * a^{2} * a^{4} = 105 * a^{7} = 105a^{7}
(ii) Length = 2p, Breadth = 4q, Height = 8r
âˆ´ Volume = Length * Breadth * Height
= 2p * 4q * 8r
= (2 * 4 * 8) * p * q * r = 64 * pqr = 64pqr
(iii) Length = xy, Breadth = 2x^{2}y, Height = 2xy^{2}
âˆ´ Volume = Length* Breadth* Height
= xy* 2x^{2}y* 2xy^{2}
= (1* 2* 2)* xy* x^{2}y* xy^{2}
= 4* x^{4}y^{4} = 4x^{4}y^{4}
(iv) Length = a, Breadth = 2b, Height = 3c
âˆ´ Volume = Length* Breadth* Height
= a* 2b* 3c
= (1* 2* 3)* a* b* c
= 6* abc = 6abc
Question 5. Obtain the product of
(i) xy, yz, zx (ii) a, â€“ a^{2}, a^{3} (iii) 2, 4y, 8y^{2}, 16y^{3} (iv) a, 2b, 3c, 6abc (v) m, â€“ mn, mnp
Solution:
(i) xy *yz *zx = (1 *1 *1) *x *y *y *z *z *x
= 1 *(x^{2} *y^{2} *z^{2}) = x^{2}y^{2}z^{2}
(ii) a *(â€“a^{2}) *a^{3} = [1 *(â€“1) *1] *a *a^{2} *a^{3}
= (â€“1) *a^{6} = â€“a^{6}
(iii) 2 *4y *8y^{2} *16y^{3} = (2 *4 *8 *16) *y *y^{2} *y^{3}
= 1024 *y^{6} = 1024y^{6}
(iv) a *2b *3c *6abc = (1 *2 *3 *6) *a *b *c *abc
= 36 *a^{2}b^{2}c^{2} = 36a^{2}b^{2}c^{2}
(v) m *(â€“mn) *mnp = [1 *(â€“1) *1] *m *mn *mnp
= (â€“1)m^{3}n^{2}p = â€“m^{3}n^{2}p
Multiplying a Monomial by a Binomial
Question: Find the product
(i) 2x(3x + 5xy) (ii) a^{2}(2ab â€“ 5c)
Solution:
(i) 2x(3x + 5xy) = 2x * 3x + 2x * 5xy
= (2 * 3) * x * x + (2 * 5) * x * xy
= 6 * x^{2} + 10 * x^{2}y
= 6x^{2} + 10x^{2}y
(ii) a^{2}(2ab â€“ 5c) = a^{2} * 2ab + a^{2} * (â€“5c)
= (1 * 2) * a^{2} * ab + [1 * (â€“5)] * a^{2} * c
= 2 * a^{3}b + (â€“5) * a^{2}c
= 2a^{3}b â€“ 5a^{2}c
Question: Find the product: (4p2 + 5p + 7) * 3p
Solution:
(4p^{2} + 5p + 7) * 3p = (4p2 * 3p) + (5p * 3p) + (7 * 3p)
= [(4 * 3) * p^{2} * p] + [(5 * 3) * p * p] + (7 * 3) * p
= 12 * p^{3} + 15 * p^{2} + 21 * p
= 12p^{3} + 15p^{2} + 21p
Example 1. Simplify: 2x(x â€“ 1) + 5 and find its value at x = â€“1
Solution:
2x(x â€“ 1) + 5 = [2x * x] + [2x * (â€“1)] + 5
= [(2 * 1) * x * x] + [2 * (â€“1) * x] + 5
= [2 * x^{2}] + [â€“2 * x] + 5
= 2x^{2} â€“ 2x + 5
For x = â€“1, 2x^{2} â€“ 2x + 5 = 2(â€“1)^{2} â€“ 2(â€“1) + 5
= (2 * 1) â€“ (â€“2) + 5
= 2 + 2 + 5 = 9