answer:

step-by-step explanation:

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) = xm+n

thus, (x³ × x⁵) = x⁸, (x⁶ + x⁴) = x6+4 = x10, 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 -3x²y³

solution:

(6xy) × (-3x²y³)

= {6 × (-3)} × {xy × x²y³}

= -18x1+2 y1+3

= -18x³y⁴.

(ii) 7ab², -4a²b and -5abc

solution:

(7ab²) × (-4a²b) × (-5abc)

= {7 × (-4) × (-5)} × {ab² × a²b × abc}

= 140 a1+2+1 b2+1+1 c

= 140a⁴b⁴c.

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) 5a²b² × (3a² - 4ab + 6b²)

solution:

5a²b² × (3a² - 4ab + 6b²)

= (5a²b²) × (3a²) + (5a²b²) × (-4ab) + (5a²b²) × (6b²)

= 15a⁴b² - 20a³b³ + 30a²b⁴.

(ii) (-3x²y) × (4x²y - 3xy² + 4x - 5y)

solution:

(-3x²y) × (4x²y - 3xy² + 4x - 5y)

= (-3x²y) × (4x²y) + (-3x²y) × (-3xy²) + (-3x²y) × (4x) + (-3x²y) × (-5y)

= -12x⁴y² + 9x³y³ - 12x³y + 15x²y².

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

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.

column wise multiplication

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).

(ii) multiply (3x² + y²) by (2x² + 3y²)

solution:

horizontal method,

= 3x² (2x² + 3y²) + y² (2x² + 3y²)

= (6x⁴ + 9x²y²) + (2x²y² + 3y⁴)

= 6x⁴ + 9x²y² + 2x²y² + 3y⁴

= 6x⁴ + 11x²y² + 3y⁴

column methods,

3x² + y²

× (2x² + 3y³)

6x⁴ + 2x²y² ⇐ multiplication by 2x² .

+ 9x²y² + 3y⁴ ⇐ multiplication by 3y³.

6x⁴ + 11x²y² + 3y⁴ ⇐ multiplication by (2x² + 3y³).

iv. multiplication by polynomial

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

(i) multiply (5x² – 6x + 9) by (2x -3)

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

(ii) multiply (2x² – 5x + 4) by (x² + 7x – 8)

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.

(iii) multiply (2x³ – 5x² – x + 7) by (3 - 2x + 4x²)

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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