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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