Question

which polynomial is factored completely?
$4(4x^2-1)$
$2x(y^3-4y^2+5y)$
$3x(9x^2+1)$
$5x^2-17x+14$

Explanation:

Step1: Check Option 1

$4(4x^2 - 1)$ can be factored further using the difference of squares formula $a^2 - b^2=(a-b)(a+b)$. Here, $4x^2 -1=(2x-1)(2x+1)$, so it is not fully factored.

Step2: Check Option 2

$2x(y^3 - 4y^2 + 5y)$ has a common factor $y$ in the parentheses. We can factor out $y$ to get $2xy(y^2 - 4y + 5)$, so it is not fully factored.

Step3: Check Option 3

$3x(3x^2 + 1)$: The quadratic $3x^2 +1$ has no real roots (since $3x^2 +1=0$ gives $x^2=-\frac{1}{3}$, which has no real solutions), so it cannot be factored further over the real numbers. This polynomial is completely factored.

Step4: Check Option 4

$5x^2 -17x +14$ can be factored into $(5x-7)(x-2)$, so it is not fully factored.

Answer:

C. $3x(3x^2+1)$