## College Algebra 7th Edition

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