## College Algebra 7th Edition

$P(x)=(2x-1)(x+2)^{2}$ zeros: $-2, \displaystyle \frac{1}{2}$
Descart's rule of signs: $P(x)=2x^{3}+7x^{2}+4x-4$ ... 1 sign changes: 1 positive zeros $P(-x)=-2x^{3}+7x^{2}-4x-4$ ... $2$ sign changes: 2 or 0 negative zeros Rational Zeros Theorem: Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,4}{1,2}$ Testing with synthetic division, try 1 $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 2 &7 &4 & -4 & & & \\\hline & 2 &18 & 44 & & & \\\hline 2 &9 &22 & |\ \ 40 & & & \end{array}$ $1$ fails (not a zero) , but we now know that 1 is an upper bound (all entries in the bottom row are positive) Try 1/2 $\left.\begin{array}{l} 1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr} 2 &7 &4 & -4 & & & \\\hline & 1 &4 & -4 & & & \\\hline 2 &8 &8 & |\ \ 0 & & & \end{array}$ $P(x)=(x-\displaystyle \frac{1}{2})(2x^{2}+8x+8)$ $=(x-\displaystyle \frac{1}{2})\cdot 2(x^{2}+4x+4)$ ... recognize a perfect square $=2(x-\displaystyle \frac{1}{2})(x+2)^{2}$ $P(x)=(2x-1)(x+2)^{2}$ zeros: $-2, \displaystyle \frac{1}{2}$