## College Algebra 7th Edition

$P(x)=(2x-1)(2x+3)(3x-1)$ zeros: $-\displaystyle \frac{3}{2},\frac{1}{3},\frac{1}{2}$
Descart's rule of signs: $P(x)=12x^{3}-20x^{2}+x+3$ ... $2$ sign changes: 2 or 0 positive zeros $P(-x)=-12x^{3}-20x^{2}-x+3$ ... $1$ sign changes: 1 negative zeros Rational Zeros Theorem: Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,3}{1,2,3,4,6,12}$ Testing with synthetic division, try 1 $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 12 &-20 & 1 & 3 & & & \\\hline & 12 & 8 & 9 & & & \\\hline 12& 8 & 9 & |\ \ 12 & & & \end{array}$ $1$ is not a zero, but it is an upper bound (bottom row all positive) try 1/2 $\left.\begin{array}{l} 1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 12 &-20 & 1 & 3 & & & \\\hline & 6 & -7 & -3 & & & \\\hline 12& -14 & -6 & |\ \ 0 & & & \end{array}$ $P(x)=(x-\displaystyle \frac{1}{2})(12x^{2}-14x-6)$ $=(x-\displaystyle \frac{1}{2})\cdot 2(6x^{2}-7x-3)$ $=(2x-1)(6x^{2}-7x-3)$ ... searching for factors of $6$($-3$)=$-18$ whose sum is $-7$ ... we find $-9$ and $+2$. $6x^{2}-7x-3=6x^{2}+9x-2x-3$ $=3x(2x+3)-(2x+3)$ $=(2x+3)(3x-1)$ $P(x)=(2x-1)(2x+3)(3x-1)$ zeros: $-\displaystyle \frac{3}{2},\frac{1}{3},\frac{1}{2}$