## College Algebra 7th Edition

$P(x)=(4x-3)(2x+1)(3x+2)$ zeros: $-\displaystyle \frac{1}{2}, -\frac{2}{3}, \frac{3}{4}$
Descart's rule of signs: $P(x)=24x^{3}+10x^{2}-13x-6$ ... $1$ sign changes: 1 positive zeros $P(-x)=-24x^{3}+10x^{2}+13x-6$ ... $2$ sign changes: 2 or 0 negative zeros Rational Zeros Theorem: Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,3,6}{1,2,3,4,6,8,12,24}$ Testing with synthetic division, try 1 $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 24 &10 &-13 & -6 & & & \\\hline & 24 & 34 & 21 & & & \\\hline 24&34 & 21 & |\ \ 15 & & & \end{array}$ $1$ is not a zero, but it is an upper bound (bottom row all positive) trying 1/2, 1/3, 2/3 all fail. Next up, 3/4: $\left.\begin{array}{l} 3/4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 24 &10 &-13 & -6 & & & \\\hline & 18 & 21 & 6 & & & \\\hline 24&28 & 8 & |\ \ 0 & & & \end{array}$ $P(x)=(x-\displaystyle \frac{3}{4})(24x^{2}+28x+8)$ $=(x-\displaystyle \frac{3}{4})\cdot 4(6x^{2}+7x+2)$ $=(4x-3)(6x^{2}+7x+2)$ ... searching for factors of 6(2)=12 whose sum is 7 ... we find 3 and 4. $6x^{2}+7x+2=6x^{2}+3x+4x+2$ $=3x(2x+1)+2(2x+1)$ $=(2x+1)(3x+2)$ $P(x)=(4x-3)(2x+1)(3x+2)$ zeros: $-\displaystyle \frac{1}{2}, -\frac{2}{3}, \frac{3}{4}$