Factor completely. 39z^3+26z^2-12z-8 square

To factor the polynomial \(39z^3 + 26z^2 - 12z - 8\) completely, we can follow these steps:<br /><br />1. **Group the terms**: Group the polynomial into two pairs.<br /> \[<br /> (39z^3 + 26z^2) + (-12z - 8)<br /> \]<br /><br />2. **Factor out the greatest common factor (GCF) from each group**:<br /> \[<br /> 13z^2(3z + 2) - 4(3z + 2)<br /> \]<br /><br />3. **Factor by grouping**: Notice that \((3z + 2)\) is a common factor.<br /> \[<br /> (13z^2 - 4)(3z + 2)<br /> \]<br /><br />4. **Factor the quadratic term**: The quadratic \(13z^2 - 4\) is a difference of squares.<br /> \[<br /> 13z^2 - 4 = ( \sqrt{13}z - 2)( \sqrt{13}z + 2)<br /> \]<br /><br />5. **Combine all factors**:<br /> \[<br /> ( \sqrt{13}z - 2)( \sqrt{13}z + 2)(3z + 2)<br /> \]<br /><br />So, the completely factored form of the polynomial \(39z^3 + 26z^2 - 12z - 8\) is:<br />\[<br />( \sqrt{13}z - 2)( \sqrt{13}z + 2)(3z + 2)<br />\]