Factor completely. 12f^5-8f^4-9f^3+6f^2 square

To factor the expression \(12f^5 - 8f^4 - 9f^3 + 6f^2\) completely, we first look for the greatest common factor (GCF) of all the terms.<br /><br />The GCF of \(12f^5\), \(8f^4\), \(9f^3\), and \(6f^2\) is \(3f^2\). We factor out \(3f^2\) from each term:<br /><br />\[ 12f^5 - 8f^4 - 9f^3 + 6f^2 = 3f^2(4f^3 - \frac{8}{3}f^2 - 3f + 2) \]<br /><br />Next, we need to factor the quadratic expression inside the parentheses. We can rewrite it as:<br /><br />\[ 4f^3 - 8f^2 - 9f + 12 \]<br /><br />We can try factoring by grouping or using synthetic division, but in this case, it is more straightforward to use the quadratic formula to find the roots of the polynomial \(4f^3 - 8f^2 - 9f + 12 = 0\).<br /><br />The quadratic formula is given by:<br /><br />\[ f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />For the polynomial \(4f^3 - 8f^2 - 9f + 12\), we have \(a = 4\), \(b = -8\), and \(c = -9\). Plugging these values into the quadratic formula gives:<br /><br />\[ f = \frac{8 \pm \sqrt{64 + 144}}{8} \]<br />\[ f = \frac{8 \pm \sqrt{208}}{8} \]<br />\[ f = \frac{8 \pm 4\sqrt{13}}{8} \]<br />\[ f = 1 \pm \frac{\sqrt{13}}{2} \]<br /><br />So the roots are \(f = 1 + \frac{\sqrt{13}}{2}\) and \(f = 1 - \frac{\sqrt{13}}{2}\).<br /><br />Thus, the factored form of the original expression is:<br /><br />\[ 12f^5 - 8f^4 - 9f^3 + 6f^2 = 3f^2(4f^3 - 8f^2 - 9f + 12) \]<br /><br />Therefore, the completely factored form is:<br /><br />\[ \boxed{3f^2(4f^3 - 8f^2 - 9f + 12)} \]