Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Sure, let's go through the factoring step by step for each expression:
### 7) [tex]\( 4x^2 - 16xy + 16y^2 \)[/tex]
To factor the expression, notice that it resembles a perfect square trinomial:
[tex]\[ 4x^2 - 16xy + 16y^2 = (2x - 4y)^2 \][/tex]
Further simplifying, we get:
[tex]\[ 4(x - 2y)^2 \][/tex]
So, the factored form is:
[tex]\[ 4(x - 2y)^2 \][/tex]
### 8) [tex]\( 8x^3 - 64y^3 \)[/tex]
This expression can be factored as the difference of cubes:
[tex]\[ 8x^3 - 64y^3 = (2x)^3 - (4y)^3 \][/tex]
Using the difference of cubes formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex], where [tex]\( a = 2x \)[/tex] and [tex]\( b = 4y \)[/tex], we get:
[tex]\[ (2x - 4y)((2x)^2 + (2x)(4y) + (4y)^2) \][/tex]
Simplifying the terms inside the parentheses:
[tex]\[ (2x - 4y)(4x^2 + 8xy + 16y^2) \][/tex]
So, the factored form is:
[tex]\[ 8(x - 2y)(x^2 + 2xy + 4y^2) \][/tex]
### 9) [tex]\( 5x^2 + 8x - 4 \)[/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\( 5(-4) = -20 \)[/tex] and add to [tex]\( 8 \)[/tex]. These numbers are [tex]\( 10 \)[/tex] and [tex]\( -2 \)[/tex].
We split the middle term using these numbers:
[tex]\[ 5x^2 + 10x - 2x - 4 \][/tex]
Then, we group the terms and factor by grouping:
[tex]\[ 5x(x + 2) - 2(x + 2) \][/tex]
Factoring out the common term [tex]\( (x + 2) \)[/tex], we get:
[tex]\[ (x + 2)(5x - 2) \][/tex]
So, the factored form is:
[tex]\[ (x + 2)(5x - 2) \][/tex]
### 10) [tex]\( 5x^3 + 3xy + 6y^2 + 10x^2 y \)[/tex]
To factor this expression, we first look for common factors in the terms:
Group terms in pairs:
[tex]\[ (5x^3 + 10x^2y) + (3xy + 6y^2) \][/tex]
From the first group, we can factor out [tex]\( 5x^2 \)[/tex]:
[tex]\[ 5x^2(x + 2y) \][/tex]
From the second group, we can factor out [tex]\( 3y \)[/tex]:
[tex]\[ 3y(x + 2y) \][/tex]
Combining these groups, we get:
[tex]\[ 5x^2(x + 2y) + 3y(x + 2y) \][/tex]
We can factor out the common binomial [tex]\( (x + 2y) \)[/tex]:
[tex]\[ (x + 2y)(5x^2 + 3y) \][/tex]
So, the factored form is:
[tex]\[ (x + 2y)(5x^2 + 3y) \][/tex]
In summary, the factored forms of the given expressions are:
7) [tex]\( 4(x - 2y)^2 \)[/tex]
8) [tex]\( 8(x - 2y)(x^2 + 2xy + 4y^2) \)[/tex]
9) [tex]\( (x + 2)(5x - 2) \)[/tex]
10) [tex]\( (x + 2y)(5x^2 + 3y) \)[/tex]
### 7) [tex]\( 4x^2 - 16xy + 16y^2 \)[/tex]
To factor the expression, notice that it resembles a perfect square trinomial:
[tex]\[ 4x^2 - 16xy + 16y^2 = (2x - 4y)^2 \][/tex]
Further simplifying, we get:
[tex]\[ 4(x - 2y)^2 \][/tex]
So, the factored form is:
[tex]\[ 4(x - 2y)^2 \][/tex]
### 8) [tex]\( 8x^3 - 64y^3 \)[/tex]
This expression can be factored as the difference of cubes:
[tex]\[ 8x^3 - 64y^3 = (2x)^3 - (4y)^3 \][/tex]
Using the difference of cubes formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex], where [tex]\( a = 2x \)[/tex] and [tex]\( b = 4y \)[/tex], we get:
[tex]\[ (2x - 4y)((2x)^2 + (2x)(4y) + (4y)^2) \][/tex]
Simplifying the terms inside the parentheses:
[tex]\[ (2x - 4y)(4x^2 + 8xy + 16y^2) \][/tex]
So, the factored form is:
[tex]\[ 8(x - 2y)(x^2 + 2xy + 4y^2) \][/tex]
### 9) [tex]\( 5x^2 + 8x - 4 \)[/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\( 5(-4) = -20 \)[/tex] and add to [tex]\( 8 \)[/tex]. These numbers are [tex]\( 10 \)[/tex] and [tex]\( -2 \)[/tex].
We split the middle term using these numbers:
[tex]\[ 5x^2 + 10x - 2x - 4 \][/tex]
Then, we group the terms and factor by grouping:
[tex]\[ 5x(x + 2) - 2(x + 2) \][/tex]
Factoring out the common term [tex]\( (x + 2) \)[/tex], we get:
[tex]\[ (x + 2)(5x - 2) \][/tex]
So, the factored form is:
[tex]\[ (x + 2)(5x - 2) \][/tex]
### 10) [tex]\( 5x^3 + 3xy + 6y^2 + 10x^2 y \)[/tex]
To factor this expression, we first look for common factors in the terms:
Group terms in pairs:
[tex]\[ (5x^3 + 10x^2y) + (3xy + 6y^2) \][/tex]
From the first group, we can factor out [tex]\( 5x^2 \)[/tex]:
[tex]\[ 5x^2(x + 2y) \][/tex]
From the second group, we can factor out [tex]\( 3y \)[/tex]:
[tex]\[ 3y(x + 2y) \][/tex]
Combining these groups, we get:
[tex]\[ 5x^2(x + 2y) + 3y(x + 2y) \][/tex]
We can factor out the common binomial [tex]\( (x + 2y) \)[/tex]:
[tex]\[ (x + 2y)(5x^2 + 3y) \][/tex]
So, the factored form is:
[tex]\[ (x + 2y)(5x^2 + 3y) \][/tex]
In summary, the factored forms of the given expressions are:
7) [tex]\( 4(x - 2y)^2 \)[/tex]
8) [tex]\( 8(x - 2y)(x^2 + 2xy + 4y^2) \)[/tex]
9) [tex]\( (x + 2)(5x - 2) \)[/tex]
10) [tex]\( (x + 2y)(5x^2 + 3y) \)[/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
