We appreciate your visit to Solve by factoring completely 7 tex x 3 125 0 tex 8 tex x 4 4x 2 5 0 tex 9 tex x 3 2x. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Sure! Let's solve each of these equations by factoring completely:
7) [tex]\( x^3 - 125 = 0 \)[/tex]
This equation is a difference of cubes, which can be factored using the formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = x \)[/tex] and [tex]\( b = 5 \)[/tex]. So, the factorization is:
[tex]\[ x^3 - 125 = (x - 5)(x^2 + 5x + 25) \][/tex]
8) [tex]\( x^4 - 4x^2 - 5 = 0 \)[/tex]
To factor this, consider substituting [tex]\( y = x^2 \)[/tex]. This transforms the equation into:
[tex]\[ y^2 - 4y - 5 = 0 \][/tex]
This can be factored as:
[tex]\[ (y - 5)(y + 1) = 0 \][/tex]
Substituting back [tex]\( x^2 \)[/tex] for [tex]\( y \)[/tex], the factors are:
[tex]\[ (x^2 - 5)(x^2 + 1) \][/tex]
9) [tex]\( x^3 + 2x^2 + 2x + 4 = 0 \)[/tex]
We can factor this by grouping:
- Group the first two terms and the last two terms: [tex]\( x^2(x + 2) + 2(x + 2) \)[/tex].
- Notice the common factor [tex]\( (x + 2) \)[/tex].
- Factor out [tex]\( (x + 2) \)[/tex]:
[tex]\[ (x + 2)(x^2 + 2) \][/tex]
10) [tex]\( x^4 - 4 = 0 \)[/tex]
This is a difference of squares, which factors as:
[tex]\[ (x^2 - 2)(x^2 + 2) \][/tex]
11) [tex]\( x^4 + 2x^2 - 35 = 0 \)[/tex]
Again let [tex]\( y = x^2 \)[/tex]. This makes the equation:
[tex]\[ y^2 + 2y - 35 = 0 \][/tex]
Factor as:
[tex]\[ (y - 5)(y + 7) = 0 \][/tex]
Substituting back [tex]\( x^2 \)[/tex] for [tex]\( y \)[/tex], we get:
[tex]\[ (x^2 - 5)(x^2 + 7) \][/tex]
12) [tex]\( x^5 - 9x^3 + 18x = 0 \)[/tex]
First, factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x(x^4 - 9x^2 + 18) \][/tex]
Consider [tex]\( x^2 \)[/tex] as another substitution [tex]\( z = x^2 \)[/tex]:
[tex]\[ z^2 - 9z + 18 = 0 \][/tex]
Factor this quadratic:
[tex]\[ (z - 6)(z - 3) \][/tex]
Substituting [tex]\( x^2 \)[/tex] back for [tex]\( z \)[/tex]:
[tex]\[ x(x^2 - 6)(x^2 - 3) \][/tex]
These are the factored forms of each equation. If you have any more questions or need further clarification, feel free to ask!
7) [tex]\( x^3 - 125 = 0 \)[/tex]
This equation is a difference of cubes, which can be factored using the formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\( a = x \)[/tex] and [tex]\( b = 5 \)[/tex]. So, the factorization is:
[tex]\[ x^3 - 125 = (x - 5)(x^2 + 5x + 25) \][/tex]
8) [tex]\( x^4 - 4x^2 - 5 = 0 \)[/tex]
To factor this, consider substituting [tex]\( y = x^2 \)[/tex]. This transforms the equation into:
[tex]\[ y^2 - 4y - 5 = 0 \][/tex]
This can be factored as:
[tex]\[ (y - 5)(y + 1) = 0 \][/tex]
Substituting back [tex]\( x^2 \)[/tex] for [tex]\( y \)[/tex], the factors are:
[tex]\[ (x^2 - 5)(x^2 + 1) \][/tex]
9) [tex]\( x^3 + 2x^2 + 2x + 4 = 0 \)[/tex]
We can factor this by grouping:
- Group the first two terms and the last two terms: [tex]\( x^2(x + 2) + 2(x + 2) \)[/tex].
- Notice the common factor [tex]\( (x + 2) \)[/tex].
- Factor out [tex]\( (x + 2) \)[/tex]:
[tex]\[ (x + 2)(x^2 + 2) \][/tex]
10) [tex]\( x^4 - 4 = 0 \)[/tex]
This is a difference of squares, which factors as:
[tex]\[ (x^2 - 2)(x^2 + 2) \][/tex]
11) [tex]\( x^4 + 2x^2 - 35 = 0 \)[/tex]
Again let [tex]\( y = x^2 \)[/tex]. This makes the equation:
[tex]\[ y^2 + 2y - 35 = 0 \][/tex]
Factor as:
[tex]\[ (y - 5)(y + 7) = 0 \][/tex]
Substituting back [tex]\( x^2 \)[/tex] for [tex]\( y \)[/tex], we get:
[tex]\[ (x^2 - 5)(x^2 + 7) \][/tex]
12) [tex]\( x^5 - 9x^3 + 18x = 0 \)[/tex]
First, factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x(x^4 - 9x^2 + 18) \][/tex]
Consider [tex]\( x^2 \)[/tex] as another substitution [tex]\( z = x^2 \)[/tex]:
[tex]\[ z^2 - 9z + 18 = 0 \][/tex]
Factor this quadratic:
[tex]\[ (z - 6)(z - 3) \][/tex]
Substituting [tex]\( x^2 \)[/tex] back for [tex]\( z \)[/tex]:
[tex]\[ x(x^2 - 6)(x^2 - 3) \][/tex]
These are the factored forms of each equation. If you have any more questions or need further clarification, feel free to ask!
Thanks for taking the time to read Solve by factoring completely 7 tex x 3 125 0 tex 8 tex x 4 4x 2 5 0 tex 9 tex x 3 2x. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
