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Which polynomial is factored completely?

A. [tex]\(4(4x^4 - 1)\)[/tex]

B. [tex]\(2x(y^3 - 4y^2 + 5y)\)[/tex]

C. [tex]\(3x(9x^2 + 1)\)[/tex]

D. [tex]\(5x^2 - 17x + 14\)[/tex]

Sagot :

GinnyAnsweridn
To determine which polynomial is factored completely, we need to analyze each of the given polynomials and see if they can be factored further.

1. First Polynomial: [tex]\(4(4x^4 - 1)\)[/tex]
- [tex]\(4x^4 - 1\)[/tex] is a difference of squares: [tex]\( (2x^2)^2 - 1^2 \)[/tex]
- The difference of squares can be factored as [tex]\( (2x^2 - 1)(2x^2 + 1) \)[/tex]
- Thus, [tex]\(4(4x^4 - 1)\)[/tex] can be factored further into [tex]\( 4(2x^2 - 1)(2x^2 + 1) \)[/tex].

2. Second Polynomial: [tex]\(2x(y^3 - 4y^2 + 5y)\)[/tex]
- The expression inside the parentheses, [tex]\( y^3 - 4y^2 + 5y \)[/tex], has a common factor of [tex]\( y \)[/tex]:
- Factoring out [tex]\( y \)[/tex] gives [tex]\( y(y^2 - 4y + 5) \)[/tex]
- Therefore, the polynomial can be factored as [tex]\( 2xy(y^2 - 4y + 5) \)[/tex].

3. Third Polynomial: [tex]\(3x(9x^2 + 1)\)[/tex]
- The expression inside the parentheses, [tex]\( 9x^2 + 1 \)[/tex], does not factor further using real numbers.
- The term [tex]\(9x^2 + 1\)[/tex] is already in its simplest form as a sum of squares cannot generally be factored over the reals.

4. Fourth Polynomial: [tex]\( 5x^2 - 17x + 14 \)[/tex]
- This is a quadratic polynomial and might be factored into the form [tex]\((ax + b)(cx + d)\)[/tex].
- We need to find two numbers that multiply to [tex]\(5 \times 14 = 70\)[/tex] and add up to [tex]\(-17\)[/tex]. These numbers are [tex]\(-7\)[/tex] and [tex]\(-10\)[/tex].
- The polynomial can be factored as [tex]\((5x - 7)(x - 2)\)[/tex].

Now, let's summarize which polynomials are factored completely:

- [tex]\(4(4x^4 - 1)\)[/tex] can be factored further, hence not completely factored.
- [tex]\(2x(y^3 - 4y^2 + 5y)\)[/tex] can be factored further, hence not completely factored.
- [tex]\(3x(9x^2 + 1)\)[/tex] cannot be factored further, indicating it is already completely factored.
- [tex]\(5x^2 - 17x + 14\)[/tex] can be factored into [tex]\((5x - 7)(x - 2)\)[/tex], so the given form wasn't in factored form originally.

Therefore, the polynomial [tex]\(3x(9x^2 + 1)\)[/tex] is factored completely.
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