In multiplication of algebraic expression before taking up the product of algebraic expressions, let us look at two simple rules.
(i) The product of two factors with like signs is positive, and the product of two factors with unlike signs is negative.
(ii) if x is a variable and m, n are positive integers, then
(xᵐ × xⁿ) = x\(^{m + n}\)
Thus, (x³ × x⁵) = x⁸, (x⁶ + x⁴) = x\(^{6 + 4}\) = x\(^{10}\), etc.
Rule:
Product of two monomials = (product of their numerical coefficients) × (product of their variable parts)
Solution:
(6xy) × (3x²y³)
= {6 × (3)} × {xy × x²y³}
= 18x\(^{1 + 2}\) y\(^{1 + 3}\)
= 18x³y⁴.
Solution:
(7ab²) × (4a²b) × (5abc)
= {7 × (4) × (5)} × {ab² × a²b × abc}
= 140 a\(^{1 + 2 + 1}\) b\(^{2 + 1 + 1}\) c
= 140a⁴b⁴c.
Rule:
Multiply each term of the polynomial by the monomial, using the distributive law a × (b + c) = a × b + a × c.
(i) 5a²b² × (3a²  4ab + 6b²)
Solution:
5a²b² × (3a²  4ab + 6b²)
= (5a²b²) × (3a²) + (5a²b²) × (4ab) + (5a²b²) × (6b²)
= 15a⁴b²  20a³b³ + 30a²b⁴.
Solution:
(3x²y) × (4x²y  3xy² + 4x  5y)
= (3x²y) × (4x²y) + (3x²y) × (3xy²) + (3x²y) × (4x) + (3x²y) × (5y)
= 12x⁴y² + 9x³y³  12x³y + 15x²y².
Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.
(a + b) × (c + d)
= a × (c + d) + b × (c + d)
= (a × c + a × d) + (b × c + b × d)
= ac + ad + bc + bd
Note: This method is known as the horizontal method.
(i) Multiply (3x + 5y) and (5x  7y).
Solution:
(3x + 5y) × (5x  7y)
= 3x × (5x  7y) + 5y × (5x  7y)
= (3x × 5x  3x × 7y) + (5y × 5x  5y × 7y)
= (15x²  21xy) + (25xy  35y²)
= 15x²  21xy + 25xy  35y²
= 15x² + 4xy  35y².
The multiplication can be performed column wise as shown below.
3x + 5y
× (5x  7y)
_____________
15x² + 25xy ⇐ multiplication by 5x.
 21xy  35y² ⇐ multiplication by 7y.
__________________
15x² + 4xy  35y² ⇐ multiplication by (5x  7y).
__________________
Solution:
= 3x² (2x² + 3y²) + y² (2x² + 3y²)
= (6x⁴ + 9x²y²) + (2x²y² + 3y⁴)
= 6x⁴ + 9x²y² + 2x²y² + 3y⁴
= 6x⁴ + 11x²y² + 3y⁴
3x² + y²
× (2x² + 3y³)
_____________
6x⁴ + 2x²y² ⇐ multiplication by 2x² .
+ 9x²y² + 3y⁴ ⇐ multiplication by 3y³.
___________________
6x⁴ + 11x²y² + 3y⁴ ⇐ multiplication by (2x² + 3y³).
___________________
We may extend the above result for two polynomials, as shown below.
5x² – 6x + 9
× (2x  3)
____________________
10x³  12x² + 18x ⇐ multiplication by 2x.
 15x² + 18x  27 ⇐ multiplication by 3.
______________________
10x³ – 27x² + 36x  27 ⇐ multiplication by (2x  3).
______________________
Therefore, (5x² – 6x + 9) by (2x  3) is 10x³ – 27x² + 36x – 27
Solution:
By column method
2x² – 5x + 4
× (x² + 7x – 8)
___________________________
2x⁴ – 5x³ + 4x² ⇐ multiplication by x².
+ 14x³  35x² + 28x ⇐ multiplication by 7x.
 16x² + 40x  32 ⇐ multiplication by 8.
___________________________
2x⁴ – 9x³  47x² + 68x  32 ⇐ multiplication by (x² + 7x  8).
___________________________
Therefore, (2x² – 5x + 4) by (x² + 7x – 8) is 2x⁴ – 9x³  47x² + 68x – 32.
Solution:
Arranging the terms of the given polynomials in descending power of x and then multiplying,
2x³ – 5x² – x + 7
× (3  2x + 4x²)
_________________________________
8x⁵  20x⁴ – 4x³ + 28x² ⇐ multiplication by 3.
 4x⁴ + 10x³ + 2x² – 14x ⇐ multiplication by 2x.
+ 6x³ – 15x²  3x + 21 ⇐ multiplication by 4x².
_________________________________
8x⁵ – 24x⁴ + 12x³ + 15x² – 17x + 21 ⇐ multiplication by (3  2x + 4x²).
_________________________________
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