Factor completely, or state that the polynomial is prime. Check the factorization using multiplication or a graphing utility. y^6-256y^2 Select the correct choice below and fill in any answer boxes within your choice. A. y^6-256y^2=square B. The polynomial is prime

To factor the polynomial \( y^{6} - 256y^{2} \), we can start by factoring out the greatest common factor (GCF) from both terms. The GCF here is \( y^{2} \).<br /><br />\[ <br />y^{6} - 256y^{2} = y^{2}(y^{4} - 256) <br />\]<br /><br />Next, we need to factor \( y^{4} - 256 \). Notice that this is a difference of squares:<br /><br />\[ <br />y^{4} - 256 = (y^{2})^{2} - (16)^{2} <br />\]<br /><br />Using the difference of squares formula, \( a^{2} - b^{2} = (a - b)(a + b) \), we can factor it as follows:<br /><br />\[ <br />y^{4} - 256 = (y^{2} - 16)(y^{2} + 16) <br />\]<br /><br />Now, let's check if \( y^{2} - 16 \) and \( y^{2} + 16 \) can be factored further.<br /><br />1. \( y^{2} - 16 \) is another difference of squares:<br /> \[<br /> y^{2} - 16 = (y - 4)(y + 4)<br /> \]<br /><br />2. \( y^{2} + 16 \) cannot be factored further over the real numbers because it is a sum of squares.<br /><br />Putting it all together, the complete factorization of the original polynomial is:<br /><br />\[ <br />y^{6} - 256y^{2} = y^{2}(y - 4)(y + 4)(y^{2} + 16) <br />\]<br /><br />Thus, the correct choice is:<br /><br />A. \( y^{6}-256 y^{2}= y^{2}(y - 4)(y + 4)(y^{2} + 16) \)<br /><br />To verify, you can expand the factors back to ensure they multiply to the original polynomial or use a graphing utility to confirm that the roots correspond to the factors.