which polynomial is factored completely?\n4(4x^4 - 1)\n2x(y^3 - 4y^2 + 5y)\n3x(9x^2 + 1)\n5x^2 - 17x + 14

Explanation:

Step1: Analyze the first option

The expression $4(4x^{4}-1)$ can be further factored. Since $4x^{4}-1=(2x^{2} + 1)(2x^{2}-1)$ and $2x^{2}-1=( \sqrt{2}x + 1)(\sqrt{2}x - 1)$. So it is not completely factored.

Step2: Analyze the second option

The expression $2x(y^{3}-4y^{2}+5y)$ can be further factored by taking out the common - factor $y$ from the terms inside the parentheses. We get $2xy(y^{2}-4y + 5)$. So it is not completely factored.

Step3: Analyze the third option

The expression $3x(9x^{2}+1)$ cannot be factored further over the real - numbers. The sum of squares $a^{2}+b^{2}$ (here $a = 3x$ and $b = 1$) has no real - number factors of the form $(mx + n)(px+q)$.

Step4: Analyze the fourth option

The expression $5x^{2}-17x + 14$ is a quadratic expression. For a quadratic $ax^{2}+bx + c$ ($a = 5$, $b=-17$, $c = 14$), we calculate the discriminant $\Delta=b^{2}-4ac=(-17)^{2}-4\times5\times14=289 - 280=9$. Then we can factor it as $5x^{2}-17x + 14=5x^{2}-10x-7x + 14=5x(x - 2)-7(x - 2)=(5x - 7)(x - 2)$. So it is not completely factored.

Answer:

$3x(9x^{2}+1)$