## College Algebra 7th Edition

$P(x)=(x-1)(2x-1)(2x+3)$ zeros: $-\displaystyle \frac{3}{2}, \frac{1}{2}, 1$
Descart's rule of signs: $P(x)=4x^{3}-7x+3$ ... $2$ sign changes: 2 or 0 positive zeros $P(-x)=-4x^{3}+7x+3$ ... 1 sign changes: 1 negative zeros Rational Zeros Theorem: Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,3}{1,2,4}$ Testing with synthetic division, ( missing powers: 0 in top row) try 1 $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 4 &0 &-7 & 3 & & & \\\hline & 4 & 4 & -3 & & & \\\hline 4 &4 &-3 & |\ \ 0 & & & \end{array}$ $P(x)=(x-1)(4x^{2}+4x-3)$ ... search for two factors of 4(-3)=-12 whose sum is 4 ... find 6 and -2 $4x^{2}+4x-3 =4x^{2}-2x+6x-3$ $=2x(2x-1)+3(2x-1)$ $=(2x-1)(2x+3)$ $P(x)=(x-1)(2x-1)(2x+3)$ zeros: $-\displaystyle \frac{3}{2}, \frac{1}{2}, 1$