## Multiplication of Algebraic Expression

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

(x^{m} × x^{n}) = x^{ (m + n)}. Thus, (x

^{3} × x

^{5}) = x

^{8}, (x

^{6} + x

^{4}) = x

^{(6 + 4)} = x

^{10}, etc.

### I. Multiplication of Two Monomials

**Rule: ** Product of two monomials = (product of their numerical coefficients) × (product of their variable parts)

### Find the product of:

### (i) 6xy and -3x^{2}y^{3}

**Solution: **
(6xy) × (-3x

^{2}y

^{3})

= {6 × (-3)} × {xy × x

^{2}y

^{3}}

= -18x

^{1 + 2} y

^{1 + 3}
= -18x

^{3}y

^{4}.

### (ii) 7ab^{2}, -4a^{2}b and -5abc

**Solution: **
(7ab

^{2}) × (-4a

^{2}b) × (-5abc)

= {7 × (-4) × (-5)} × {ab

^{2} × a

^{2}b × abc}

= 140 a

^{1 + 2 + 1}
b

^{2 + 1 + 1}
c

= 140a

^{4}b

^{4}c.

### II. Multiplication of a Polynomial by a Monomial

**Rule: **
Multiply each term of the polynomial by the monomial, using the distributive law a × (b + c) = a × b + a × c.

### Find each of the following products:

### (i) 5a^{2}b^{2} × (3a^{2} - 4ab + 6b^{2})

**Solution: **
5a

^{2}b

^{2} × (3a

^{2} - 4ab + 6b

^{2})

= (5a

^{2}b

^{2}) × (3a

^{2}) + (5a

^{2}b

^{2}) × (-4ab) + (5a

^{2}b

^{2}) × (6b

^{2})

= 15a

^{4}b

^{2} - 20a

^{3}b

^{3} + 30a

^{2}b

^{4}.

### (ii) (-3x^{2}y) × (4x^{2}y - 3xy^{2} + 4x - 5y)

**Solution: **
(-3x

^{2}y) × (4x

^{2}y - 3xy

^{2} + 4x - 5y)

= (-3x

^{2}y) × (4x

^{2}y) + (-3x

^{2}y) × (-3xy

^{2}) + (-3x

^{2}y) × (4x) + (-3x

^{2}y) × (-5y)

= -12x

^{4}y

^{2} + 9x

^{3}y

^{3} - 12x

^{3}y + 15x

^{2}y

^{2}.

### III. Multiplication of Two Binomials

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

*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)

= (15x

^{2} - 21xy) + (25xy - 35y

^{2})

= 15x

^{2} - 21xy + 25xy - 35y

^{2}
= 15x

^{2} + 4xy - 35y

^{2}.

*Column wise multiplication *

The multiplication can be performed column wise as shown below.

3x + 5y

× (5x - 7y)

_____________

15x

^{2} + 25xy

⇐ multiplication by 5x.
- 21xy - 35y^{2} ⇐ multiplication by -7y.

__________________

15x^{2} + 4xy - 35y^{2} ⇐ multiplication by (5x - 7y).

__________________

### (ii) Multiply (3x^{2} + y^{2}) by (2x^{2} + 3y^{2})

**Solution: **
*Horizontal method, *

= 3x

^{2} (2x

^{2} + 3y

^{2}) + y

^{2} (2x

^{2} + 3y

^{2})

= (6x

^{4} + 9x

^{2}y

^{2}) + (2x

^{2}y

^{2} + 3y

^{4})

= 6x

^{4} + 9x

^{2}y

^{2} + 2x

^{2}y

^{2} + 3y

^{4}
= 6x

^{4} + 11x

^{2}y

^{2} + 3y

^{4}
*Column methods, *

3x^{2} + y^{2}

× (2x^{2} + 3y^{3})

_____________

6x^{4} + 2x^{2}y^{2} ⇐ multiplication by 2x^{2} .

+ 9x^{2}y^{2} + 3y^{4} ⇐ multiplication by 3y^{3}.

___________________

6x^{4} + 11x^{2}y^{2} + 3y^{4} ⇐ multiplication by (2x^{2} + 3y^{3}).

___________________

### IV. Multiplication by Polynomial

We may extend the above result for two polynomials, as shown below.

### (i) Multiply (5x^{2} – 6x + 9) by (2x -3)

5x^{2} – 6x + 9

× (2x - 3)

____________________

10x^{3} - 12x^{2} + 18x ⇐ multiplication by 2x.

- 15x^{2} + 18x - 27 ⇐ multiplication by -3.

______________________

10x^{3} – 27x^{2} + 36x – 27 ⇐ multiplication by (2x - 3).

______________________

Therefore, (5x^{2} – 6x + 9) by (2x - 3) is 10x^{3} – 27x^{2} + 36x – 27

### (ii) Multiply (2x^{2} – 5x + 4) by (x^{2} + 7x – 8)

**Solution: **
By column method

2x

^{2} – 5x + 4

× (x

^{2} + 7x – 8)

___________________________

2x

^{4} – 5x

^{3} + 4x

^{2} ⇐ multiplication by x^{2}.
+ 14x

^{3} - 35x

^{2} + 28x

⇐ multiplication by 7x.
- 16x

^{2} + 40x - 32

⇐ multiplication by -8.
___________________________

2x

^{4} – 9x

^{3} - 47x

^{2} + 68x - 32

⇐ multiplication by (x^{2} + 7x - 8).
___________________________

Therefore, (2x

^{2} – 5x + 4) by (x

^{2} + 7x – 8) is 2x

^{4} – 9x

^{3} - 47x

^{2} + 68x – 32.

### (iii) Multiply (2x^{3} – 5x^{2} – x + 7) by (3 - 2x + 4x^{2})

**Solution: **
Arranging the terms of the given polynomials in descending power of x and then multiplying,

2x

^{3} – 5x

^{2} – x + 7

× (3 - 2x + 4x

^{2})

_________________________________

8x

^{5} - 20x

^{4} – 4x

^{3} + 28x

^{2} ⇐ multiplication by 3.
- 4x

^{4} + 10x

^{3} + 2x

^{2} – 14x

⇐ multiplication by -2x.
+ 6x

^{3} – 15x

^{2} - 3x + 21

⇐ multiplication by 4x^{2}.
_________________________________

8x

^{5} – 24x

^{4} + 12x

^{3} + 15x

^{2} – 17x + 21

⇐ multiplication by (3 - 2x + 4x^{2}).
_________________________________

● **Algebraic Expression**
●

**Algebraic Expression**
●

**Addition of Algebraic Expressions**
●

**Subtraction of Algebraic Expressions**
●

**Multiplication of Algebraic Expression**
●

**Division of Algebraic Expressions**
8th Grade Math Practice

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