## College Algebra 7th Edition

$P(x)=(x-2)(x-3)(x+2)(3x-1)$ zeros: $-2, \displaystyle \frac{1}{3}, 2, 3$
Descart's rule of signs: $P(x)=3x^{4}-10x^{3}-9x^{2}+40x-12$ ... 3 sign changes: 3 or 1 positive zeros $P(-x)=3x^{4}+10x^{3}-9x^{2}-40x-12$ ... 1 sign changes: 1 negative zeros Rational Zeros Theorem: Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,3,4,6,12}{1,3}$ Testing with synthetic division, 1 fails, try 2 $\left.\begin{array}{l} 2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 3 &-10 &-9 & 40 & -12 & & \\\hline & 6 &-8 & -34 & 12 & & \\\hline 3 &-4 &-17 & 6 & |\ \ 0 & & \end{array}$ $P(x)=(x-2)(3x^{3}-4x^{2}-17x+6)$ Testing with synthetic division again, 2 fails, try 3: $\left.\begin{array}{l} 3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 3 &-4 & -17 & 6 & & & \\\hline & 9 &15 & -6 & & & \\\hline 3 &5 &-2 & |\ \ 0 & & & \end{array}$ $P(x)=(x-2)(x-3)(3x^{2}+5x-2)$ We factor the trinomial by searching for factors of 3(-2)=-6 whose sum is 5 ... we find 6 and -1 $3x^{2}+5x^{2}-2=3x^{2}+6x-x-2$ $=3x(x+2)-(x+2)=(x+2)(3x-1)$ $P(x)=(x-2)(x-3)(x+2)(3x-1)$ zeros: $-2, \displaystyle \frac{1}{3}, 2, 3$