Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(C) has accurate answers prepared by experts at BYJU’S, with the intention to improve grasping abilities of students. Multiplication of algebraic expressions is the major concept discussed under this exercise. Students who aim to score well in the annual exam are suggested to solve the textbook questions, using these solutions. To speed up the problem solving abilities, students can make use of Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(C) PDF, from the links which are provided, with a free download option.

## Selina Solutions Concise Mathematics Class 6 Chapter 19: Fundamental Operations Exercise 19(C) Download PDF

### Access other exercises of Selina Solutions Concise Mathematics Class 6 Chapter 19: Fundamental Operations

### Access Selina Solutions Concise Mathematics Class 6 Chapter 19: Fundamental Operations Exercise 19(C)

Exercise 19(C)

**1. Fill in the blanks:**

**(i) 6 Ã— 3 = â€¦â€¦â€¦. and 6x Ã— 3x = â€¦â€¦â€¦â€¦**

**(ii) 6 Ã— 3 = â€¦â€¦… and 6x ^{2} Ã— 3x^{3} = â€¦â€¦â€¦â€¦**

**(iii) 5 Ã— 4 = â€¦â€¦â€¦. and 5x Ã— 4y = â€¦â€¦â€¦â€¦**

**(iv) 4 Ã— 7 = â€¦â€¦â€¦â€¦. and 4ax Ã— 7x = â€¦â€¦â€¦.**

**(v) 6 Ã— 2 = â€¦â€¦â€¦â€¦. and 6xy Ã— 2xy = â€¦â€¦….**

**Solution:**

(i) 6 Ã— 3 = 18

Hence,

6x Ã— 3x = 6 Ã— 3 Ã— x Ã— x

We get,

= 18 **Ã— **x^{2}

= 18x^{2}

Therefore, 6 Ã— 3 = 18 and 6x Ã— 3x = 18x^{2}

(ii) 6 Ã— 3 = 18

Hence,

6x^{2} Ã— 3x^{3} = 6 Ã— 3 Ã— x^{2 + 3}

= 18 **Ã— **x^{5}

= 18x^{5}

Therefore, 6 Ã— 3 = 18 and 6x^{2} Ã— 3x^{3 }= 18x^{5}

(iii) 5 Ã— 4 = 20 and 5x Ã— 4y = 5 Ã— 4 Ã— x Ã— y

= 20xy

Therefore, 5 Ã— 4 = 20 and 5x Ã— 4y = 20xy

(iv) 4 **Ã— 7 = **28

Hence,

4ax Ã— 7x = 4 **Ã— 7 Ã— **a Ã— x Ã— x

= 28 Ã— a Ã— x^{2}

= 28ax^{2}

Therefore, 4 Ã— 7 = 28 and 4ax Ã— 7x = 28ax^{2}

(v) 6 **Ã— **2 = 12

Hence,

6xy Ã— 2xy = 6 Ã— 2 Ã— x^{1 + 1} Ã— y^{1 + 1}

= 12 Ã— x^{2} Ã— y^{2}

= 12x^{2}y^{2}

Therefore, 6 Ã— 2 = 12 and 6xy Ã— 2xy = 12x^{2}y^{2}

**2. Fill in the blanks:**

**(i) 4x Ã— 6x Ã— 2 = â€¦â€¦â€¦**

**(ii) 3ab Ã— 6ax = â€¦â€¦â€¦â€¦**

**(iii) x Ã— 2x ^{2} Ã— 3x^{3} = â€¦â€¦â€¦**

**(iv) 5 Ã— 5a ^{3} = â€¦â€¦â€¦â€¦**

**(v) 6 Ã— 6x ^{2} Ã— 6x^{2}y^{2} = â€¦â€¦â€¦**

**Solution:**

(i) 4x Ã— 6x Ã— 2 = 4 Ã— 6 Ã— 2 Ã— x Ã— x

= 48 Ã— x^{2}

We get,

= 48x^{2}

Hence, 4x Ã— 6x Ã— 2 = 48x^{2}

(ii) 3ab **Ã— **6ax = 3 Ã— 6 Ã— a Ã— a Ã— b Ã— x

= 18 Ã— a^{2} Ã— b Ã— x

We get,

= 18a^{2}bx

Hence, 3ab Ã— 6ax = 18a^{2}bx

(iii) x **Ã— **2x^{2} Ã— 3x^{3} = 2 Ã— 3 Ã— x Ã— x^{2} Ã— x^{3}

= 6 Ã— x^{1 + 2 + 3}

= 6 Ã— x^{6}

= 6x^{6}

Hence, x Ã— 2x^{2} Ã— 3x^{3} = 6x^{6}

(iv) 5 Ã— 5a^{3} = 5 Ã— 5 Ã— a^{3}

= 25 Ã— a^{3}

We get,

= 25a^{3}

Hence, 5 Ã— 5a^{3} = 25a^{3}

(v) 6 **Ã— **6x^{2} Ã— 6x^{2}y^{2} = 6 Ã— 6 Ã— 6 Ã— x^{2} Ã— x^{2} Ã— y^{2}

= 216 Ã— x^{2+ 2} Ã— y^{2}

= 216 Ã— x^{4} Ã— y^{2}

We get,

= 216x^{4}y^{2}

Hence, 6 Ã— 6x^{2} Ã— 6x^{2}y^{2} = 216x^{4}y^{2}

**3. Find the value of:**

**(i) 3x ^{3} Ã— 5x^{4}**

**(ii) 5a ^{2} Ã— 7a^{7}**

**(iii) 3abc Ã— 6ac ^{3}**

**(iv) a ^{2}b^{2} Ã— 5a^{3}b^{4}**

**(v) 2x ^{2}y^{3} Ã— 5x^{3}y^{4}**

**Solution:**

(i) 3x^{3} Ã— 5x^{4}

3x^{3} Ã— 5x^{4} = 3 Ã— 5 Ã— x^{3} Ã— x^{4}

= 15 Ã— x^{3 + 4}

We get,

= 15 Ã— x^{7}

= 15x^{7}

Hence, the value of 3x^{3} Ã— 5x^{4} is 15x^{7}

(ii) 5a^{2} Ã— 7a^{7}

5a^{2} Ã— 7a^{7} = 5 Ã— 7 Ã— a^{2} Ã— a^{7}

= 35 Ã— a^{2 }^{+ 7}

= 35 Ã— a^{9}

We get,

= 35a^{9}

Hence, the value of 5a^{2} Ã— 7a^{7} is 35a^{9}

(iii) 3abc Ã— 6ac^{3}

3abc Ã— 6ac^{3} = 3 Ã— 6 Ã— a Ã— a Ã— b Ã— c Ã— c^{3}

=18 Ã— a^{1+ 1} Ã— b Ã— c^{1+3}

= 18 Ã— a^{2} Ã— b Ã— c^{4}

We get,

= 18a^{2}bc^{4}

Hence, the value of 3abc Ã— 6ac^{3} is 18a^{2}bc^{4}

(iv) a^{2}b^{2} Ã— 5a^{3}b^{4}

a^{2}b^{2} Ã— 5a^{3}b^{4} = 5 Ã— a^{2} Ã— a^{3} Ã— b^{2} Ã— b^{4}

= 5 Ã— a^{2+3} Ã— b^{2+4}

= 5 Ã— a^{5} Ã— b^{6}

We get,

= 5a^{5}b^{6}

Hence, the value of a^{2}b^{2} Ã— 5a^{3}b^{4} is 5a^{5}b^{6}

(v) 2x^{2}y^{3} Ã— 5x^{3}y^{4}

2x^{2}y^{3} Ã— 5x^{3}y^{4} = 2 Ã— 5 Ã— x^{2} Ã— x^{3} Ã— y^{3} Ã— y^{4}

= 10 Ã— x^{2+3} Ã— y^{3+4}

We get,

= 10 Ã— x^{5} Ã— y^{7}

= 10x^{5}y^{7}

Hence, the value of 2x^{2}y^{3} Ã— 5x^{3}y^{4} is 10x^{5}y^{7}

**4. Multiply:**

**(i) a + b by ab**

**(ii) 3ab â€“ 4b by 3ab**

**(iii) 2xy â€“ 5by by 4bx**

**(iv) 4x + 2y by 3xy**

**(v) 1 + 4x by x**

**Solution:**

(i) a + b by ab

The multiplication of a + b by ab is calculated as,

(a + b) Ã— ab = a Ã— ab + b Ã— ab

= a^{1+1}b + ab^{1+1}

We get,

= a^{2}b + ab^{2}

Hence, (a + b) by ab = a^{2}b + ab^{2}

(ii) 3ab â€“ 4b by 3ab

The multiplication of 3ab â€“ 4b by 3ab is calculated as,

(3ab â€“ 4b)** **Ã— 3ab = 3ab Ã— 3ab â€“ 4b Ã— 3ab

= 9a^{1+1}b^{1+1} â€“ 12ab^{1+1}

We get,

= 9a^{2}b^{2} â€“ 12ab^{2}

Therefore, (3ab â€“ 4b) by 3ab = 9a^{2}b^{2} â€“ 12ab^{2}

(iii) 2xy â€“ 5by by 4bx

The multiplication of 2xy â€“ 5by by 4bx is calculated as,

(2xy â€“ 5by) Ã— 4bx = 2xy Ã— 4bx â€“ 5by Ã— 4bx

= 8bx^{1+1}y â€“ 20b^{1+1}xy

We get,

= 8bx^{2}y â€“ 20b^{2}xy

Therefore, (2xy â€“ 5by) by 4bx = 8bx^{2}y â€“ 20b^{2}xy

(iv) 4x + 2y by 3xy

The multiplication of 4x + 2y by 3xy is calculated as,

(4x + 2y) Ã— 3xy = 4x Ã— 3xy + 2y Ã— 3xy

On simplification, we get

= 12x^{1+1}y + 6xy^{1+1}

= 12x^{2}y + 6xy^{2}

Therefore, (4x + 2y) by 3xy = 12x^{2}y + 6xy^{2}

(v) 1 + 4x by x

The multiplication of (1 + 4x) by x is calculated as,

(1 + 4x) Ã— x = 1 Ã— x + 4x Ã— x

On simplification, we get

= x + 4x^{1+1}

= x + 4x^{2}

Therefore, (1 + 4x) by x = x + 4x^{2}

**5. Multiply:**

**(i) â€“ x + y â€“ z and â€“ 2x**

**(ii) xy â€“ yz and x ^{2}yz^{2}**

**(iii) 2xyz + 3xy and â€“ 2y ^{2}z**

**(iv) â€“ 3xy ^{2} + 4x^{2}y and â€“ xy**

**(v) 4xy and â€“ x ^{2}y â€“ 3x^{2} y^{2}**

**Solution:**

(i) â€“ x + y â€“ z and â€“ 2x

The multiplication of the given expression is calculated as,

(- x + y â€“ z) Ã— – 2x = – x Ã— – 2x + y Ã— – 2x â€“ z Ã— – 2x

On further calculation, we get

= 2x^{1+1} â€“ 2xy + 2xz

= 2x^{2} â€“ 2xy + 2xz

Hence, the multiplication of (- x + y â€“ z) and â€“ 2x is 2x^{2} â€“ 2xy + 2xz

(ii) xy â€“ yz and x^{2}yz^{2}

The multiplication of the given expression is calculated as,

(xy â€“ yz) Ã— (x^{2}yz^{2}) = xy Ã— x^{2}yz^{2} â€“ yz Ã— x^{2}yz^{2}

We get,

= x^{1+2}y^{1+1}z^{2} â€“ x^{2}y^{1+1}z^{1+2}

= x^{3}y^{2}z^{2} â€“ x^{2}y^{2}z^{3}

Hence, the multiplication of (xy â€“ yz) and x^{2}yz^{2} = x^{3}y^{2}z^{2} â€“ x^{2}y^{2}z^{3}

(iii) 2xyz + 3xy and â€“ 2y^{2}z

The multiplication of the given expression is calculated as,

(2xyz + 3xy) Ã— – 2y^{2}z = 2xyz Ã— – 2y^{2}z + 3xy Ã— – 2y^{2}z

On further calculation, we get

= – 4xy^{1+2}z^{1+1} â€“ 6xy^{1+2}z

= – 4xy^{3}z^{2} â€“ 6xy^{3}z

Hence, the multiplication of 2xyz + 3xy and â€“ 2y^{2}z = – 4xy^{3}z^{2} â€“ 6xy^{3}z

(iv) â€“ 3xy^{2} + 4x^{2}y and â€“ xy

The multiplication of the given expression is calculated as,

(- 3xy^{2} + 4x^{2}y) Ã— – xy = 3x^{1+1}y^{2+1} â€“ 4x^{2+1}y^{1+1}

On calculation, we get

= 3x^{2}y^{3} â€“ 4x^{3}y^{2}

Hence, the multiplication of â€“ 3xy^{2} + 4x^{2}y and â€“ xy = 3x^{2}y^{3} â€“ 4x^{3}y^{2}

(v) 4xy and â€“ x^{2}y â€“ 3x^{2} y^{2}

The multiplication of the given expression is calculated as,

(- x^{2}y â€“ 3x^{2}y^{2}) Ã— 4xy = – x^{2}y Ã— 4xy â€“ 3x^{2}y^{2} Ã— 4xy

On further calculation, we get

= – 4x^{2+1}y^{1+1} â€“ 12x^{2+1}y^{2+1}

= – 4x^{3}y^{2} â€“ 12x^{3}y^{3}

Hence, the multiplication of 4xy and â€“ x^{2}y â€“ 3x^{2} y^{2} = – 4x^{3}y^{2} â€“ 12x^{3}y^{3}

**6. Multiply:**

**(i) 3a + 4b â€“ 5c and 3a**

**(ii) â€“ 5xy and â€“ xy ^{2} â€“ 6x^{2}y**

**Solution:**

(i) 3a + 4b â€“ 5c and 3a

The multiplication of the given expression is calculated as,

(3a + 4b â€“ 5c) Ã— 3a = 3a Ã— 3a + 4b Ã— 3a â€“ 5c Ã— 3a

On further calculation, we get

= 9a^{1+1} + 12ab â€“ 15ac

= 9a^{2} + 12ab â€“ 15ac

Therefore, the multiplication of 3a + 4b â€“ 5c and 3a = 9a^{2} + 12ab â€“ 15ac

(ii) â€“ 5xy and â€“ xy^{2} â€“ 6x^{2}y

The multiplication of the given expression is calculated as,

– 5xy Ã— (- xy^{2} â€“ 6x^{2}y) = – 5xy Ã— – xy^{2} â€“ 5xy Ã— – 6x^{2}y

On further calculation, we get

= 5x^{1+1}y^{1+2} + 30x^{1+2}y^{1+1}

= 5x^{2}y^{3} + 30x^{3}y^{2}

Therefore, the multiplication of â€“ 5xy and â€“ xy^{2} â€“ 6x^{2}y = 5x^{2}y^{3} + 30x^{3}y^{2}

**7. Multiply:**

**(i) x + 2 and x + 10**

**(ii) x + 5 and x â€“ 3**

**(iii) x â€“ 5 and x + 3**

**(iv) x â€“ 5 and x â€“ 3**

**(v) 2x + y and x + 3y**

**Solution:**

(i) x + 2 and x + 10

The given expression is calculated as follows

(x + 2) Ã— (x + 10) = x Ã— (x + 10) + 2 Ã— (x + 10)

We get,

= x^{2} + 10x + 2x + 20

= x^{2} + 12x + 20

Hence, the multiplication of (x + 2) and (x + 10) = x^{2} + 12x + 20

(ii) x + 5 and x â€“ 3

The given expression is calculated as follows

(x + 5) Ã— (x â€“ 3) = x Ã— (x â€“ 3) + 5 Ã— (x â€“ 3)

On simplification, we get

= x^{2} â€“ 3x + 5x â€“ 15

= x^{2} + 2x â€“ 15

Hence, the multiplication of (x + 5) and (x â€“ 3) = x^{2} + 2x â€“ 15

(iii) x â€“ 5 and x + 3

The given expression is calculated as follows

(x â€“ 5) Ã— (x + 3) = x Ã— (x + 3) â€“ 5 Ã— (x + 3)

On further calculation, we get

= x^{2} + 3x â€“ 5x â€“ 15

= x^{2} â€“ 2x – 15

Hence, the multiplication of (x â€“ 5) and (x + 3) = x^{2} â€“ 2x â€“ 15

(iv) x â€“ 5 and x â€“ 3

The given expression is calculated as,

(x â€“ 5) Ã— (x â€“ 3) = x Ã— (x â€“ 3) â€“ 5 Ã— (x â€“ 3)

On further calculation, we get

= x^{2} â€“ 3x â€“ 5x + 15

= x^{2} â€“ 8x + 15

Hence, the multiplication of (x â€“ 5) and (x â€“ 3) = x^{2} â€“ 8x + 15

(v) 2x + y and x + 3y

The given expression is calculated as,

(2x + y) Ã— (x + 3y) = 2x Ã— (x + 3y) + y Ã— (x + 3y)

On simplification, we get

= 2x^{2} + 6xy + xy + 3y^{2}

= 2x^{2} + 7xy + 3y^{2}

Hence, the multiplication of (2x + y) and (x + 3y) = 2x^{2} + 7xy + 3y^{2}

**8. Multiply:**

**(i) 3abc and â€“ 5a ^{2}b^{2}c**

**(ii) x â€“ y + z and -2x**

**(iii) 2x â€“ 3y â€“ 5z and -2y**

**(iv) â€“ 8xyz + 10 x ^{2}yz^{3} and xyz**

**(v) xyz and â€“ 13xy ^{2}z + 15x^{2}yz â€“ 6xyz^{2}**

**Solution:**

(i) 3abc and â€“ 5a^{2}b^{2}c

The given expression is calculated as follows,

3abc Ã— – 5a^{2}b^{2}c = 3 Ã— – 5 Ã— a Ã— a^{2} Ã— b Ã— b^{2} Ã— c Ã— c

On further calculation, we get

= – 15 Ã— a^{1+2} Ã— b^{1+2} Ã— c^{1+1}

= – 15 Ã— a^{3} Ã— b^{3} Ã— c^{2}

= – 15a^{3}b^{3}c^{2}

Therefore, the multiplication of 3abc and â€“ 5a^{2}b^{2}c = – 15a^{3}b^{3}c^{2}

(ii) x â€“ y + z and -2x

The given expression is calculated as follows,

(x â€“ y + z) Ã— – 2x = x Ã— – 2x â€“ y Ã— – 2x + z Ã— – 2x

On simplification, we get

= – 2x^{1+1} + 2xy â€“ 2xz

= – 2x^{2} + 2xy â€“ 2xz

Therefore, the multiplication of x â€“ y + z and -2x = – 2x^{2} + 2xy â€“ 2xz

(iii) 2x â€“ 3y â€“ 5z and -2y

The given expression is calculated as follows,

(2x â€“ 3y â€“ 5z) Ã— – 2y = 2x Ã— – 2y â€“ 3y Ã— – 2y â€“ 5z Ã— – 2y

On further calculation, we get

= – 4xy + 6y^{1+1} + 10yz

= – 4xy + 6y^{2} + 10yz

Therefore, the multiplication of 2x â€“ 3y â€“ 5z and -2y = – 4xy + 6y^{2} + 10yz

(iv) â€“ 8xyz + 10 x^{2}yz^{3} and xyz

The given expression is calculated as follows,

(- 8xyz + 10x^{2}yz^{3}) Ã— xyz = – 8xyz Ã— xyz + 10x^{2}yz^{3} Ã— xyz

On further calculation, we get

= – 8x^{1+1}y^{1+1}z^{1+1} + 10x^{2+1}y^{1+1}z^{3+1}

= – 8x^{2}y^{2}z^{2} + 10x^{3}y^{2}z^{4}

Therefore, the multiplication of â€“ 8xyz + 10 x^{2}yz^{3} and xyz = – 8x^{2}y^{2}z^{2} + 10x^{3}y^{2}z^{4}

(v) xyz and â€“ 13xy^{2}z + 15x^{2}yz â€“ 6xyz^{2}

The given expression is calculated as follows,

xyz Ã— (- 13xy^{2}z + 15x^{2}yz â€“ 6xyz^{2}) = xyz Ã— – 13xy^{2}z + xyz Ã—15x^{2}yz â€“ xyz Ã— 6xyz^{2}

On simplification, we get

= – 13x^{1+1}y^{1+2}z^{1+1} + 15x^{1+2}y^{1+1}z^{1+1} â€“ 6x^{1+1}y^{1+1}z^{1+2}

We get,

= – 13x^{2}y^{3}z^{2} + 15x^{3}y^{2}z^{2} â€“ 6x^{2}y^{2}z^{3}

Therefore, the multiplication of xyz and â€“ 13xy^{2}z + 15x^{2}yz â€“ 6xyz^{2} = – 13x^{2}y^{3}z^{2} + 15x^{3}y^{2}z^{2} â€“ 6x^{2}y^{2}z^{3}

**9. Find the product of:**

**(i) xy â€“ ab and xy + ab**

**(ii) 2abc â€“ 3xy and 2abc + 3xy**

**(iii) a + b â€“ c and 2a â€“ 3b**

**(iv) 5x â€“ 6y â€“ 7z and 2x + 3y**

**(v) 5x â€“ 6y â€“ 7z and 2x + 3y + z**

**Solution:**

(i) xy â€“ ab and xy + ab

The product of the given expression is calculated as,

(xy â€“ ab) Ã— (xy + ab) = xy Ã— (xy + ab) â€“ ab Ã— (xy + ab)

On simplification, we get

= xy Ã— xy + xy Ã— ab â€“ ab Ã— xy â€“ ab Ã— ab

= x^{2}y^{2} + abxy â€“ abxy â€“ a^{2}b^{2}

= x^{2}y^{2} â€“ a^{2}b^{2}

Hence, the product of (xy â€“ ab) and (xy + ab) = x^{2}y^{2} â€“ a^{2}b^{2}

(ii) 2abc â€“ 3xy and 2abc + 3xy

The product of the given expression is calculated as,

(2abc â€“ 3xy) Ã— (2abc + 3xy)

= 2abc Ã— (2abc + 3xy) â€“ 3xy Ã— (2abc + 3xy)

We get,

= 2abc Ã— 2abc + 2abc Ã— 3xy â€“ 3xy Ã— 2abc â€“ 3xy Ã— 3xy

= 4a^{2}b^{2}c^{2} + 6abcxy â€“ 6abcxy â€“ 9x^{2}y^{2}

= 4a^{2}b^{2}c^{2} â€“ 9x^{2}y^{2}

Hence, the product of 2abc â€“ 3xy and 2abc + 3xy = 4a^{2}b^{2}c^{2} â€“ 9x^{2}y^{2}

(iii) a + b â€“ c and 2a â€“ 3b

The product of the given expression is calculated as,

(a + b â€“ c) Ã— (2a â€“ 3b)

= a Ã— (2a â€“ 3b) + b Ã— (2a â€“ 3b) â€“ c Ã— (2a â€“ 3b)

= a Ã— 2a â€“ a Ã— 3b + b Ã— 2a â€“ b Ã— 3b â€“ c Ã— 2a + c Ã— 3b

= 2a^{1+1} â€“ 3ab + 2ab â€“ 3b^{1+1} â€“ 2ac + 3bc

We get,

= 2a^{2} â€“ ab â€“ 3b^{2} â€“ 2ac + 3bc

Hence, the product of a + b â€“ c and 2a â€“ 3b = 2a^{2} – ab â€“ 3b^{2} â€“ 2ac + 3bc

(iv) 5x â€“ 6y â€“ 7z and 2x + 3y

The product of the given expression is calculated as,

(5x â€“ 6y â€“ 7z) Ã— (2x + 3y)

= (5x â€“ 6y â€“ 7z) Ã— 2x + (5x â€“ 6y â€“ 7z) Ã— 3y

= 5x Ã— 2x â€“ 6y Ã— 2x â€“ 7z Ã— 2x + 5x Ã— 3y â€“ 6y Ã— 3y â€“ 7z Ã— 3y

We get,

= 10x^{2} â€“ 12xy â€“ 14xz + 15xy â€“ 18y^{2} â€“ 21yz

= 10x^{2} + 3xy â€“ 14xz â€“ 18y^{2} â€“ 21yz

Hence, the product of 5x â€“ 6y â€“ 7z and 2x + 3y = 10x^{2} + 3xy â€“ 14xz â€“ 18y^{2} â€“ 21yz

(v) 5x â€“ 6y â€“ 7z and 2x + 3y + z

The product of the given expression is calculated as,

(5x â€“ 6y â€“ 7z) Ã— (2x + 3y + z)

= (5x â€“ 6y â€“ 7z) Ã— 2x + (5x â€“ 6y â€“ 7z) Ã— 3y + (5x â€“ 6y â€“ 7z) Ã— z

= 5x Ã— 2x â€“ 6y Ã— 2x â€“ 7z Ã— 2x + 5x Ã— 3y â€“ 6y Ã— 3y â€“ 7z Ã— 3y + 5x Ã— z â€“ 6y Ã— z â€“ 7z Ã— z

We get,

= 10x^{2} â€“ 12xy â€“ 14xz + 15xy â€“ 18y^{2} â€“ 21yz + 5xz â€“ 6yz â€“ 7z^{2}

= 10x^{2} â€“ 12xy + 15xy â€“ 14xz + 5xz â€“ 18y^{2} â€“ 21yz â€“ 6yz â€“ 7z^{2}

= 10x^{2} + 3xy â€“ 9xz â€“ 18y^{2} â€“ 27yz â€“ 7z^{2}

Hence, the product of 5x â€“ 6y â€“ 7z and 2x + 3y + z = 10x^{2} + 3xy â€“ 9xz â€“ 18y^{2} â€“ 27yz â€“ 7z^{2}