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

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(3x²−6x+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(3x²−6x+2) ⇒ −12x² + 24x − 8

Page 8 Exercise 7 Answer

Given: −1(3x²−6x+2)

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

We can write

−1(3x²−6x+2) = −(3x²−6x+2)

Further solving using the distributive principle we get

−1(3x²−6x+2) = −3x² + 6x − 2

Hence using the Distributive principle to solve −1(3x²−6x+2) we get −3x² + 6x − 2.

Page 8 Exercise 8 Answer

Given:(3x²+6x−2)(−10)

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

We can write

(3x²−60x+20)(−10) = −(30x²−60x+20)

Further solving using the Distributive principle we get

(3x²−60x+20)(−10) = −30x² + 60x − 20

Hence using Distributive principle to solve (3x²−60x+20)(−10) we get −30x² + 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 = 2x² − 7x

Hence using the Distributive principle to solve(2x−7)x we get 2x² − 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 = 3y² + 5y

Hence using Distributive principle to solve (3y+5)y we get 3y² + 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(2y²+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(2y²+3x−4) = 2xy² + 3x² − 4x

Hence using Distributive principle to solve x(2y²+3x−4) we get 2xy² + 3x² − 4x

Page 8 Exercise 7 Answer

Given:(2y²+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

(2y²+3x−4)xy = 2xy³ + 3x²y − 4xy

Hence using the Distributive principle to solve (2y²+3x−4)xy we get 2xy³ + 3x²y − 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) = 20x² − 35x

Hence using Distributive principle to solve 5x(4x−7) we get 20x² − 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) = 12a² + 6a

Hence using the Distributive principle to solve 3a(4a+2) we get 12a² + 6a.

Page 8 Exercise 3 Answer

Given: 4x(x²−5)

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

4x(x²−5) = 4x³ − 20x

Hence using Distributive principle to solve 4x(x²−5) we get 4x³ − 20x

Page 8 Exercise 4 Answer

Given: (2x+1)2x²

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)2x² = 4x³ + 2x²

Hence using the Distributive principle to solve (2x+1)2x² we get 4x³ + 2x².

Page 8 Exercise 5 Answer

Given:(3y−6)(−5y)

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 (3y−6)(−5y) = −15y² + 30y

Hence using Distributive principle to solve (3y−6)(−5y) we get −15y² + 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 21x² + 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 12a² − 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 + 20y² − 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 + 8xy² − 16xy.

Page 8 Exercise 10 Answer

We are given the expression (7x²−5x−6)6x².

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 42x⁴ − 30x³ − 36x².

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 + 12b² − 4bc.

Page 8 Exercise 12 Answer

We are given the expression 7x²y(3x²y+2xy²+x³).

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

Given: 7x²y(3x²y+2xy²+x³)

on applying the distributive principle, we get

7x²y(3x²y+2xy²+x³)=7x²y.3x²y+7x²y.2xy²+7x²y.x³
=21x⁴y²+14x³y³+14x³y³+7x⁵y

Hence, using the Distributive principle for the given multiplication problem 7x²y(3x²y+2xy²+x³), we get the result 21x⁴y² + 14x³y³ + 7x⁵y.

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 = 6x² + 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 = 35n² + 20n