## Mc Graw Hill Key To Algebra Book 4 Polynomials 1st Edition Chapter 5 The Distributive Principle

**Page 8 Exercise 1 Answer**

Given: 3(2x−5)

To find:

Simplify using the distributive principle.

Let us solve the equation using the distributive Principle.

The given equation is :

**The given equation is: **

⇒ 3(2x-5)

Use distributive principle

∴ 3(2x-5)=3x2x-3×5

∴ 3x2x-3×5=6x-15

Using distributive Principle.

3(2x−5) ⇒ 6x − 15

**Page 8 Exercise 3 Answer**

Given:

⇒ (3a+4b)2

To find:

Simplify using the distributive principle.

Let us solve the equation using the distributive Principle.

The given equation is :

Given Equation is

⇒ 2(3a+4b)

use distributive principle

∴2 (3a+4b)=2x3a+2x4b

⇒ 6a+8b

Using distributive Principle.

2(3a+4b) ⇒ 6a+8b

**Page 8 Exercise 5 Answer**

Given:

⇒ 4(3x−y+5)

To find:

Simplify using the distributive principle.

Let us solve the equation using the distributive Principle.

The given equation is :

⇒ 4(3x-y+5)

use distributive principle

∴ 4x3x-4xy+4×5

⇒ 12x-4y+20

Using distributive Principle.

4(3x−y+5) ⇒ 12x − 4y + 20

**Page 8 Exercise 7 Answer**

Given:

⇒ 6(3x−y+5)

**To find:**

Simplify using the distributive principle

Let us solve the equation using the distributive Principle.

The given equation is:

⇒ 6(3x-y+5)

use distributive principle

∴ 6x3x-6xy+6×5

⇒ 18x-6y+30

Using distributive Principle

6(3x−y+5) ⇒ 18x − 6y + 30

**Page 8 Exercise 1 Answer**

Given:

⇒ −5(2x−4)

To find:

Simplify using the distributive principle

Let us solve the equation using the distributive Principle.

The given equation is:

⇒ -5(2x-4)

Use distributive principle

∴ -5(2x-4)

Using distributive Principle

∴ -5(2x-4) ⇒-5x2x-(-5)x4

⇒ -5(2x-4)⇒-10x+20

Using distributive Principle

−5(2x−4) ⇒ −10x + 20

**Page 8 Exercise 2 Answer**

Given:

⇒ −4(3y+5)

To find:

Simplify using the distributive principle

Let us solve the equation using the distributive Principle.

Given equation is

⇒ -4(3y+5)

use distributive principle

∴ -4(3y+5)⇒-4x3y+(-4)x5

⇒ -4(3y+5)⇒-12y-20

Using distributive Principle

−4(3y+5) ⇒ −12y − 20

**Page 8 Exercise 4 Answer**

Given:

⇒ (a+x)−8

To find:

Simplify using the distributive principle

Let us solve the equation using the distributive Principle.

The given equation is:

⇒ (a+x)-8

use distributive principle

∴ (a+x)-8⇒ 1xa+1×x-8

∴ (a+x)-8⇒a+x-8

Using distributive Principle

(a+x)−8 ⇒ a + x − 8

**Page 8 Exercise 5 Answer**

Given:

⇒ −4(3x^{2}−6x+2)

To find:

Simplify using the distributive principle

Let us solve the equation using the distributive Principle.

The given equation is:

⇒ -4(3×2-6x+2)

use distributive principle

∴ -4(3×2-6x+2)

∴ \(-4\left(3 x^2-6 x+2\right)=-4 \times 3 x^2+4 \times 6 x-4 \times 2\)

∴\(-4\left(3 x^2-6 x+2\right)=-12 x^2+24 x-8\)

Using distributive Principle

−4(3x^{2}−6x+2) ⇒ −12x^{2} + 24x − 8

**Page 8 Exercise 7 Answer**

Given: −1(3x^{2}−6x+2)

**To find:** We have to use the distributive principle to multiply.

We can write

−1(3x^{2}−6x+2) = −(3x^{2}−6x+2)

Further solving using the distributive principle we get

−1(3x^{2}−6x+2) = −3x^{2} + 6x − 2

Hence using the Distributive principle to solve −1(3x^{2}−6x+2) we get −3x^{2} + 6x − 2.

**Page 8 Exercise 8 Answer**

Given:(3x^{2}+6x−2)(−10)

To find: We have to use the distributive principle to multiply.

We can write

(3x^{2}−60x+20)(−10) = −(30x^{2}−60x+20)

Further solving using the Distributive principle we get

(3x^{2}−60x+20)(−10) = −30x^{2} + 60x − 20

Hence using Distributive principle to solve (3x^{2}−60x+20)(−10) we get −30x^{2} + 60x − 20

**Page 8 Exercise 1 Answer**

Given: (2x−7)x

To find: We have to use the distributive principle to multiply.

According to the Distributive principle, we just have to multiply the single term times each term in the polynomial

We get

(2x−7)x = 2x^{2} − 7x

Hence using the Distributive principle to solve(2x−7)x we get 2x^{2} − 7x.

**Page 8 Exercise 2 Answer**

Given:(3y+5)y

To find: We have to use the distributive principle to multiply.

According to Distributive principle, we just have to multiply the single term times each term in the polynomial

We get

(3y+5)y = 3y^{2} + 5y

Hence using Distributive principle to solve (3y+5)y we get 3y^{2} + 5y

**Page 8 Exercise 3 Answer**

Given: x(5−8y)

To find: We have to use the distributive principle to multiply.

According to the Distributive principle we just have to multiply the single term times each term in the polynomial

We get

x(5−8y) = 5x − 8xy

Hence using the Distributive principle to solve x(5−8y) we get 5x − 8xy.

**Page 8 Exercise 6 Answer**

Given: x(2y^{2}+3x−4)

To find: We have to use distributive principle to do the multiplication.

According to Distributive principle, we just have to multiply the single term times each term in the polynomial

We get

x(2y^{2}+3x−4) = 2xy^{2} + 3x^{2} − 4x

Hence using Distributive principle to solve x(2y^{2}+3x−4) we get 2xy^{2} + 3x^{2} − 4x

**Page 8 Exercise 7 Answer**

Given:(2y^{2}+3x−4)xy

To find: We have to use distributive principle to do the multiplication.

According to the Distributive principle, we just have to multiply the single term times each term in the polynomial

We get

(2y^{2}+3x−4)xy = 2xy^{3} + 3x^{2}y − 4xy

Hence using the Distributive principle to solve (2y^{2}+3x−4)xy we get 2xy^{3} + 3x^{2}y − 4xy.

**Page 8 Exercise 1 Answer**

Given: 5x(4x−7)

To find: We have to use distributive principle to do the multiplication.

According to the Distributive principle, we just have to multiply the single term times each term in the polynomial

We get

5x(4x−7) = 20x^{2} − 35x

Hence using Distributive principle to solve 5x(4x−7) we get 20x^{2} − 35x

**Page 8 Exercise 2 Answer**

Given: 3a(4a+2)

To find: We have to use distributive principle to do the multiplication.

According to the Distributive principle we just have to multiply the single term times each term in the polynomial, We get,

3a(4a+2) = 12a^{2} + 6a

Hence using the Distributive principle to solve 3a(4a+2) we get 12a^{2} + 6a.

**Page 8 Exercise 3 Answer**

Given: 4x(x^{2}−5)

To find: We have to use distributive principle to do the multiplication.

We get

4x(x^{2}−5) = 4x^{3} − 20x

Hence using Distributive principle to solve 4x(x^{2}−5) we get 4x^{3} − 20x

**Page 8 Exercise 4 Answer**

Given: (2x+1)2x^{2}

To find: We have to use distributive principle to do the multiplication.

According to the Distributive principle we just have to multiply the single term times each term in the polynomial

We get(2x+1)2x^{2} = 4x^{3} + 2x^{2}

Hence using the Distributive principle to solve (2x+1)2x^{2} we get 4x^{3} + 2x^{2}.

**Page 8 Exercise 5 Answer**

Given:(3y−6)(−5y)

To find: We have to use the distributive principle to multiply.

We get (3y−6)(−5y) = −15y^{2} + 30y

Hence using Distributive principle to solve (3y−6)(−5y) we get −15y^{2} + 30y

**Page 8 Exercise 6 Answer**

We are given the expression 7x(3x+4y).

The objective is to use the Distributive Principle for a multiplication problem.

Given: 7x(3x+4y)

on Applying the distribution Principle, we get

7x(3x+4y)=7x.3x+7x.4y

=21x²+28xy

Hence, using the Distributive principle for the given multiplication problem 7x(3x+4y), we get the result 21x^{2} + 28xy.

**Page 8 Exercise 7 Answer**

We are given the expression 3a(4a−2b+c).

The objective is to use the Distributive Principle for a multiplication problem.

Given: 3a(4a-2b+c)

On applying the distributive principle

3a(4a-2b+c)=3a.4a-3a.2b+3a.c

3a(4a-2b+c)=12a²-6ab+3ac

Hence, using the Distributive principle for the given multiplication problem 3a(4a−2b+c), we get the result 12a^{2} − 6ab + 3ac.

**Page 8 Exercise 8 Answer**

We are given the expression 5y(3x+4y−8).

The objective is to use the Distributive Principle for a multiplication problem.

Given: 5y(3x+4y-8)

On applying the distributive principle, we get

5y(3x+4y-8)=5y.3x+5y.4y-5y.8

=15xy+20y²-40y

Hence, using the Distributive principle for the given multiplication problem 5y(3x+4y−8), we get the result 15xy + 20y^{2} − 40y.

**Page 8 Exercise 9 Answer**

We are given the expression 2xy(3x+4y−8).

The objective is to use the Distributive Principle for a multiplication problem.

Given: 2xy(3x+4y-8)

on applying the distributive principle, we get

2xy(3x+4y-8)=2xy.3x+2xy.4y-32xy.8

=6x2y+8xy2-16xy

Hence, using the Distributive principle for the given multiplication problem 2xy(3x+4y−8), we get the result 6x2y + 8xy^{2} − 16xy.

**Page 8 Exercise 10 Answer**

We are given the expression (7x^{2}−5x−6)6x^{2}.

The objective is to use the Distributive Principle for a multiplication problem.

Hence, using the Distributive principle for the given multiplication problem (7×2−5x−6)6×2, we get the result 42x^{4} − 30x^{3} − 36x^{2}.

**Page 8 Exercise 11 Answer**

We are given the expression −4b(a−3b+c).

The objective is to use the Distributive Principle for a multiplication problem.

Given: 2xy(3x+4y-8)

on applying the distributive principle, we get

2xy(3x+4y-8)=2xy.3x+2xy.4y-2xy.8

=6x²y+8xy²-16xy

Hence, using the Distributive principle for the given multiplication problem −4b(a−3b+c), we get the result −4ab + 12b^{2} − 4bc.

**Page 8 Exercise 12 Answer**

We are given the expression 7x^{2}y(3x^{2}y+2xy^{2}+x^{3}).

The objective is to use the Distributive Principle for a multiplication problem.

**Given: **7x²y(3x²y+2xy²+x^{3})

on applying the distributive principle, we get

7x²y(3x²y+2xy²+x³)=7x²y.3x²y+7x²y.2xy²+7x²y.x³

=21x^{4}y²+14x³y³+14x³y³+7x^{5}y

Hence, using the Distributive principle for the given multiplication problem 7x^{2}y(3x^{2}y+2xy^{2}+x^{3}), we get the result 21x^{4}y^{2} + 14x^{3}y^{3} + 7x^{5}y.

**Page 9 Exercise 1 Answer**

We are given the equation 5(x+3) = 35.

The objective is to solve the given equation.

Given: 5(x+3)=35

on applying distributive principle, we get

5x+5.3=35

5x+15=35

5×35-15

5x=20

x= \(\frac{20}{5}\)

x=4

Hence, the solution of the given equation 5(x+3) = 35 is x = 4.

**Page 9 Exercise 3 Answer**

We are given the equation 2(3x+15) = 18.

The objective is to solve the given equation.

Given: 2(3x+15)=18

on applying the distributive principle, we get

2.3x+2.15=18

6x+30=18

6x=18-30

6x=-12

\(x=\frac{-12}{6}\)

x-2

Hence, the solution of the given equation 2(3x+15) = 18 is x = −2.

**Page 9 Exercise 4 Answer**

We are given the equation 8(x−2) = 32.

The objective is to solve the given equation.

Given: 8(x-2)=32

By applying the distributive principle, we get

5x+5.3=35

5x+15=35

5x=35-15

5x=20

x= \(\frac{20}{5}\)

x=4

Hence, the solution of the given equation 8(x−2) = 32 is x = 6.

**Page 9 Exercise 5 Answer**

We are given the equation 5(a+3) = 8a.

The objective is to solve the given equation.

Given: 2(3x+15)=18

on applying distributive principle, we get

2.3x+2.15=18

6x+30=18

6x=18-30

6x=-12

x= \(\frac{-12}{6}\)

x=-2

Hence, the solution of the given equation 5(a+3) = 8a is a = 5.

**Page 9 Exercise 6 Answer**

We are given the equation 7x = 4(x+6).

The objective is to solve the given equation.

Given: 7x=4(x+6)

On applying the distributive principle, we get

7x=4.x+4.6

7x=4x+24

7x-4x=24

3x=24

\(x=\frac{24}{3}\)

x=8

Hence, the solution of the given equation 7x = 4(x+6) is x = 8.

**Page 9 Exercise 7 Answer**

We are given the equation 4(x+5) = 3(x−6).

The objective is to solve the given equation.

Given: 4(x+5)=3(x-6)

on applying the distributive principle, we get

4x+4.5=3.-3.6

4x+20=3x-18

4x-3x=-18-20

x=-38

Hence, the solution of the given equation 4(x+5) = 3(x−6) is x = −38.

**Page 9 Exercise 8 Answer**

We are given the equation 3(2x−5)+4 = 31.

The objective is to solve the given equation.

Given: 3(2x-5)+4=31

on applying the distributive principle, we get

3.2x-3.5+4=31

6x-15=31

6x-11=31

6x=31+11

6x=42

x= \(\frac{42}{6}\)

x=7

Hence, the solution of the given equation 3(2x−5) + 4 = 31 is x = 7.

**Page 9 Exercise 9 Answer**

We are given the equation 10 = 18 + 4(3x+7).

The objective is to solve the given equation.

Given: 10=18+4(3x+7)

on applying distributive principle, we get

10=18+4.3x+4.7

10=18+12x+28

10=12x+46

10-46=12x

-36=12x

x= \(\frac{-36}{12}\)

Hence, the solution of the given equation 10 = 18 + 4(3x+7) is x = −3.

**Page 9 Exercise 10 Answer**

Given expression is 3(3x+5) = 2(6x−3)

We find the solution for a given expression

We use the distributive principle for a given expression

Given: 3(3x+5)=2(6x-3)

we use the distributive principle

3.3x+3.5=2.6x-3.2

9x+15=12x-6

12x-9x=15+6

3x=21

x=\(\frac{21}{3}\)

x=7

The solution for the given polynomial is x = 7.

**Page 10 Exercise 1 Answer**

Given the length of the rectangle = 3x

Breadth of rectangle = 2x + 5

We find the polynomial for the area of the given rectangle

We write the formula for area of the rectangle,

Given: length of rectangle =3x

breadth of rectangle= 2x+5

A=lxb

A=3x(2x+5)

we use distributive principle

A=6×2+15x

The polynomial for the area of a given rectangle is A = 6x^{2} + 15x

**Page 10 Exercise 2 Answer**

Given : Length of rectangle = 3x + 4

Breadth of rectangle = 2

We find a polynomial for the area of the given rectangle

We write the formula for the area of a rectangle,

Polynomial for area of given rectangle is A = 6x + 8

**Page 10 Exercise 4 Answer**

Given: Length of rectangle = 2r + 7

Breadth of rectangle = p

We find the polynomial for area of given rectangle

We write formula for area of rectangle,

The polynomial for area of given rectangle is 2rp + 7p

**Page 10 Exercise 6 Answer**

Given: Length of rectangle = 7n + 4

Breadth of rectangle = 5n

We find the polynomial for area of given rectangle

We write the formula for area of rectangle,

Given: length of rectangle= 3x

Breadth of rectangle= 2x+5

A=lxb

a=3x(2x+5)

we use distributive principle

A=6×2+15x

The polynomial for area of a given rectangle is A = 35n^{2} + 20n