of each equation using Descartes Rule o ) 2x^3-3x^2+5x-6=0

To determine the possible number of positive and negative real roots of the equation \(2x^3 - 3x^2 + 5x - 6 = 0\) using Descartes' Rule of Signs, we need to analyze the signs of the coefficients.<br /><br />### Positive Real Roots<br />Descartes' Rule of Signs states that the number of positive real roots of a polynomial is equal to the number of sign changes between consecutive non-zero coefficients, or less than that by an even number.<br /><br />For the polynomial \(2x^3 - 3x^2 + 5x - 6\):<br />- The coefficients are: \(2, -3, 5, -6\)<br />- The signs of the coefficients are: \(+, -, +, -\)<br /><br />Count the number of sign changes:<br />- From \(2\) to \(-3\): 1 change<br />- From \(-3\) to \(5\): 1 change<br />- From \(5\) to \(-6\): 1 change<br /><br />There are 3 sign changes, so the possible number of positive real roots is 3, 1 (3 minus an even number).<br /><br />### Negative Real Roots<br />To find the possible number of negative real roots, we substitute \(x\) with \(-x\) and then apply the same rule.<br /><br />Substitute \(x\) with \(-x\):<br />\[2(-x)^3 - 3(-x)^2 + 5(-x) - 6 = -2x^3 - 3x^2 - 5x - 6\]<br /><br />The coefficients are: \(-2, -3, -5, -6\)<br />The signs of the coefficients are: \(-, -, -, -\)<br /><br />Count the number of sign changes:<br />- There are no sign changes.<br /><br />Since there are no sign changes, there are 0 negative real roots.<br /><br />### Summary<br />- Possible number of positive real roots: 3 or 1<br />- Possible number of negative real roots: 0<br /><br />Thus, the equation \(2x^3 - 3x^2 + 5x - 6 = 0\) can have either 3 or 1 positive real roots and no negative real roots.