1.4 Polynomiyhtälön ratkaiseminen

179

a)
$x^3+2x^2+5x=0$
$x\left(x^2+2x+5\right)=0$

$x=0\ tai\ x^2+2x+5=0$

$x=0\ tai\ x=-1\pm2i$
b)

$x^3+x^2+2x-4=0$

$x=-1\pm\sqrt{3}\cdot i\ tai\ x=1$

175

$x^3-5x=-2x^2+6$
$x^3+2x^2-5x-6=0$

$\pm1{,}\ \pm2{,}\ \pm3{,}\ \pm6$

$x=2{,}\ x=-1\ tai\ x=-3$

164

a) 3
b) 1, 2, 4, 8
c) -1, 2, 4
tarkistus:

176

funktio on kasvava väleillä $-1<x<2\ ja\ 2<x$

167

a)
$x^4=3x^2$
$x^4-3x^2=0$
$x^2\left(x^2-3\right)=0$
$x^2=0\ tai\ x^2-3=0$
$x=0\ tai\ x=\sqrt{3}$
b)
$2x^3+3x^2=1$
$x^2\left(2x+3\right)=1$
$x=-1\ tai\ x=\frac{1}{2}$

162

$f\left(x\right)=\frac{3x^3-11x^2-6x+8}{x-4}$

$\begin{matrix} &3x^2+x-2\\ x-4&3x^3-11x^2-6x+8\\ &3x^3+12x^2\\ &--------\\ &x^2-6x\\ &x^2-4x\\ &--------\\ &-2x+8\\ &-2x+8\\ &--------\\ &0 \end{matrix}$

eli

$\left(x-7\right)\left(3x^2+x-2\right)=0$
$x-7=0\ tai\ 3x^2+x-2=0$
$x=7$
$tai$
$x=\frac{-1\pm\sqrt{1^2+24}}{6}=\frac{-1\pm5}{6}=-1\ tai\ \frac{2}{3}$

esim

a)
$-x^2-2x-2=0$
$x=\frac{2\pm\sqrt{4-4\cdot\left(-1\right)\cdot\left(-2\right)}}{-1\cdot2}=\frac{2\pm\sqrt{4-8}}{-2}=\frac{2\pm\sqrt{4}\cdot\sqrt{-1}}{-2}=\frac{2\pm2i}{-2}=-1\pm i$
b)
$x^4+x^2=0$

$x^2\left(x^2+1\right)=0$

$x^2=0\ tai\ x^2+1=0$

$x=0\ tai\ x=\pm i$
c)
$\left(3+2i\right)+\left(2-i\right)=5+i$
d)
$\left(3+2i\right)\cdot\left(2-i\right)=4+i$