1.2 Polymien jaollisuus

121

$\frac{2x^2+8x+8}{x+2}=\frac{2\left(x^2+4x+4\right)}{x+2}$

$x^2+4x+4$
$a=1{,}\ b=4{,}\ c=4$
$u\cdot v=4\ ja\ u+v=4$
$u=2\ ja\ v=2$
$x^2+4x+4=\left(x^2+2x\right)+\left(2x+4\right)=x\left(x+2\right)+2\left(x+2\right)=\left(x+2\right)\left(x+2\right)=\left(x+2\right)^2$

$\frac{2x^2+8x+8}{x+2}=\frac{2\left(x^2+4x+4\right)}{x+2}=\frac{2\left(x+2\right)^2}{\left(x+2\right)}=2\left(x+2\right)=2x+4$

$\frac{x^2+2x+2}{x+1}=\frac{x^2+2x+1+1}{x+1}=\frac{\left(x+1\right)^2+1}{x+1}=x+1+\frac{1}{x+1}$

122

$P\left(1\right)=3\cdot1^2-1-2=3-1-2=0$

$3x^2-x-2$

$ax^2+bx+c=\left(ax^2+ux\right)+\left(vx+c\right)$

$a=3{,}\ b=-1{,}\ c=-2$
$u\cdot v=a\cdot c\ ja\ u+v=b$
$u\cdot v=-6\ ja\ u+v=-1$
$u=2\ ja\ v=-3$

$\left(3x^2+2x\right)+\left(-3x-2\right)$

$x\left(3x+2\right)-\left(3x+2\right)=\left(x-1\right)\left(3x+2\right)$

$P\left(2\right)=3\cdot2^2-2-2=12-4=8$

ei ole

$P\left(1\right)=3\cdot1^2-1-2=0$

$3x^2-x-2$
$ax^2+bx+c=\left(ax^2+ux\right)+\left(vx+c\right)$
$a=3{,}\ b=-1{,}\ c=-2$
$u\cdot v=a\cdot c\ ja\ u+v=b$
$u\cdot v=-6\ ja\ u+v=-1$
$u=2\ ja\ v=-3$
$\left(3x^2+2x\right)+\left(-3x-2\right)$
$x\left(3x+2\right)-\left(3x+2\right)=\left(x-1\right)\left(3x+2\right)$

$\frac{\left(x-1\right)\left(3x+2\right)}{2x-2}=\frac{\left(x-1\right)\left(3x+2\right)}{2\left(x-1\right)}=\frac{3x+2}{2}$
$P\left(x\right)=\left(2x-2\right)\left(\frac{3}{2}x+1\right)$

123

$3x^2-6x+3$

$a=3{,}\ b=-6{,}\ c=3$
$u\cdot v=9\ ja\ u+v=-6$

$u=-3\ ja\ v=-3$

$\left(3x^2-3x\right)+\left(-3x+3\right)=3x\left(x-1\right)-3\left(x-1\right)=\left(3x-3\right)\left(x-1\right)$

$2x^2-7x-4$
$a=2{,}\ b=-7{,}\ c=-4$
$u\cdot v=-8\ ja\ u+v=-7$
$u=-8\ ja\ v=1$

$\left(2x^2-8x\right)+\left(x-4\right)=2x\left(x-4\right)+1\left(x-4\right)=\left(2x+1\right)\left(x-4\right)$

$x^3-2x^2+3x=x\left(x^2-2x+3\right)$

124

$\frac{4x^2-4}{x+1}=\frac{4\left(x-1\right)\left(x+1\right)}{\left(x+1\right)}=4\left(x-1\right)=4x-4$

$4x^2-4=\left(4x-4\right)\left(x+1\right)$

$\frac{x^3+x^2-12x}{x+4}=\frac{x\left(x^2+x-12\right)}{x+4}$
$x=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot\left(-12\right)}}{2\cdot1}=\frac{-1\pm\sqrt[]{49}}{2}=\frac{-1\pm7}{2}$

$x=\frac{-1+7}{2}=3$

tai

$x=\frac{-1-7}{2}=\frac{-8}{2}=-4$

$\frac{x\left(x^2+x-12\right)}{x+4}=\frac{x\left(x-3\right)\left(x+4\right)}{\left(x+4\right)}=x\left(x-3\right)$

$x^3+x^2-12x=x\left(x-3\right)\left(x+4\right)$

125
a)

$P\left(\frac{1}{2}\right)=6\cdot\left(\frac{1}{2}\right)^2-\frac{1}{2}-2=1\frac{1}{2}-\frac{1}{2}-2=-1$

ei ole jaollinen

$P\left(\frac{1}{2}\right)=4\cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^3-\frac{1}{2}\cdot\left(\frac{1}{2}\right)^2=\frac{4}{16}-\frac{1}{8}-\frac{1}{8}=\frac{4}{16}-\frac{1}{4}=\frac{4}{16}-\frac{4}{16}=0$
$\frac{4x^4-x^3-\frac{1}{2}x^2}{2x-1}=\frac{4x^4-x^3-\frac{x^2}{2}}{2x-1}=\frac{\frac{4x^42}{2}-\frac{x^32}{2}-\frac{x^2}{2}}{2x-1}=\frac{\frac{8x^4-2x^3-x^2}{2}}{2x-1}=\frac{8x^4-2x^3-x^2}{2\left(2x-1\right)}=\frac{x^2\left(8x^2-2x-1\right)}{2\left(2x-1\right)}$
$f\left(x\right)=8x^2-2x-1$
$a=8{,}\ b=-2{,}\ c=-1$
$u\cdot v=-8\ ja\ u+v=-2$
$u=2\ ja\ v=-4$
$\left(8x^2+2x\right)+\left(-4x-1\right)=2x\left(4x+1\right)-1\left(4x+1\right)=\left(2x-1\right)\left(4x+1\right)$

$\frac{x^2\left(8x^2-2x-1\right)}{2\left(2x-1\right)}=\frac{x^2\left(2x-1\right)\left(4x+1\right)}{2\left(2x-1\right)}=\frac{x^2\left(4x+1\right)}{2}$

$P\left(x\right)=\left(\frac{x^2\left(4x+1\right)}{2}\right)\cdot\left(2x-1\right)$

126

A I

B II, IV

C II, IV

D III

127

$\left(x^2+1+\frac{3}{x+3}\right)\cdot\left(x+3\right)$

$x^2+1+\frac{3}{x+3}=\frac{x^2\left(x+3\right)}{x+3}+\frac{x+3}{x+3}+\frac{3}{x+3}=\frac{x^3+3x^2+x+6}{x+3}$
$P\left(x\right)=\frac{x^3+3x^2+x+6}{x+3}\cdot\left(x+3\right)=x^3+3x^2+x+6$

$\frac{2x^3-5x^2+3x}{2x-3}=\frac{x\left(2x^2-5x+3\right)}{2x-3}$
$2x^2-5x+3$
$a=2{,}\ b=-5{,}\ c=3$
$u\cdot v=6\ ja\ u+v=-5$
$u=-2\ ja\ v=-3$
$\left(2x^2-2x\right)+\left(-3x+3\right)=2x\left(x-1\right)+-3\left(x-1\right)=\left(2x-3\right)\left(x-1\right)$
$P\left(x\right)=\frac{x\left(2x-3\right)\left(x-1\right)}{2x-3}=x\left(x-1\right)=x^2-x$

128

$x^2-12x+a_1=a_2^2+2a_2b+b^2$
$a_2=x{,}\ b=-6{,}\ b^2=a_1=36$

$9x^2+ax+4=\left(3x\right)^2+\left(2\cdot b\cdot3x\right)+4\ \ \ \ \ \left|\right|4=2^2{,}\ joten\ b=\pm2$
$\left(3x\right)^2+\left(2\cdot2\cdot3x\right)+2^2=9x^2+12x+4$
$a=\pm12$

$ax^2-8x+4=ax^2+\left(2\cdot-2\cdot ax\right)+\left(-2\right)^2=\left(2x-2\right)^2$

$\left(2x-2\right)^2=4x^2+2\cdot\left(-2\right)\cdot2x+\left(-2\right)^2=4x^2-8x+4$
$a=4$

129

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

130

$x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)$

$x^3+8x^2+x+8=x^2\left(x+8\right)+\left(x+8\right)=\left(x^2+1\right)\left(x+8\right)$

$x^3-3x^2+2x-6=x^2\left(x-3\right)+2\left(x-3\right)=\left(x^2+2\right)\left(x-3\right)$

131

$2x^2+2x-24=0$

$x=\frac{-2\pm\sqrt[]{2^2-4\cdot2\cdot\left(-24\right)}}{2\cdot2}=\frac{-2\pm14}{4}$
$x=\frac{-2+14}{4}=3$
$tai$
$x=\frac{-2-14}{4}=-4$

$P\left(2\right)=2^2-2k-k^2-1=4-2k-k^2-1=-k^2-2k+3$
$-k^2-2k+3=0$
$k=\frac{2\pm\sqrt[]{\left(-2\right)^2-4\cdot\left(-1\right)\cdot3}}{2\cdot\left(-1\right)}=\frac{2\pm4}{-2}$
$k=\frac{2+4}{-2}=-3$
$tai$
$k=\frac{2-4}{-2}=1$

132

$2x^2+ax+2a$

$2x^2+a\left(x+2\right)$
$a=0$
$2x^2+0\cdot\left(x+2\right)=2x^2=2x\cdot x$
$a=16$
$2x^2+16\left(x+2\right)=2x^2+16x+32=2\left(x^2+8x+16\right)=2\left(x+4\right)^2$

133

$\frac{3x^2+5x-2}{2x+4}=\frac{x\left(3x+5\right)-2}{2\left(x+2\right)}$

$\lim_{x\rightarrow-2}\left(\frac{x\left(3x+5\right)-2}{2\left(x+2\right)}\right)=\frac{-2\left(3\cdot\left(-2\right)+5\right)-2}{2\left(-2+2\right)}=\frac{-2\left(-6+5\right)-2}{2\cdot0}=Nollalla\ jako$

Käytetään Hopitalin sääntöä

$\lim_{x\rightarrow-2}=\frac{f'\left(x\right)}{g'\left(x\right)}=\frac{6x+5}{2}=\frac{6\cdot\left(-2\right)+5}{2}=-\frac{7}{2}$

$\frac{2x^3-2x^2-3x+3}{x-1}=\frac{x^2\left(2x-2\right)-3\left(x-1\right)}{x-1}=\frac{x^2\left(2\left(x-1\right)\right)-3\left(x-1\right)}{x-1}=\frac{2x^2\left(x-1\right)-3\left(x-1\right)}{x-1}=\frac{\left(2x^2-3\right)\left(x-1\right)}{\left(x-1\right)}=2x^3-3$

$\lim_{x\rightarrow1}=2x^3-3=2\cdot1^3-3=-1$

134

$\frac{2x^2+5x-3}{3x+a}$

$2x^2+5x-3$
$a=2{,}\ b=5{,}\ c=-3$
$u\cdot v=-6\ ja\ u+v=5$
$1{,}2{,}3{,}6$
$u=-1\ ja\ v=6$
$\left(2x^2-x\right)+\left(6x-3\right)=x\left(2x-1\right)+3\left(2x-1\right)=\left(x+3\right)\left(2x-1\right)$
$\frac{\left(x+3\right)\left(2x-1\right)}{3x+a}$
$\frac{\left(x+3\right)}{3x+a}=\frac{x+3}{3\left(x+\frac{a}{3}\right)}$
$\frac{a}{3}=3$
$a=9$

tai

$\frac{2x-1}{3x+a}=\frac{2\left(x-\frac{1}{2}\right)}{3\left(x+\frac{a}{3}\right)}$
$x-\frac{1}{2}=x+\frac{a}{3}$
$x-\frac{1}{2}=\frac{3}{2}\left(x+\frac{a}{3}\right)$
$x-\frac{1}{2}=\frac{3}{2}x+\frac{3a}{6}\ \ \ \ \ \left|\right|:\ \frac{3}{2}x=\frac{3x}{2}{,}\frac{3a}{6}=\frac{a}{2}$
$x-\frac{1}{2}=\frac{3x}{2}+\frac{a}{2}$
$x-\frac{1}{2}=\frac{3x+a}{2}$
$x=\frac{3x+a+1}{2}$
$\frac{3x+a+1}{2}-x=0$
$\frac{3x+a+1-2x}{2}=0$
$\frac{x+a+1}{2}=0$
$x+a+1=0$
$x+a=-1$

135

$2x^5-x^4-8x^3+4x^2=x^4\left(2x-1\right)-4x^2\left(2x-1\right)=\left(x^4-4x^2\right)\left(2x-1\right)=x^2\left(x^2-4\right)\left(2x-1\right)=x\cdot x\left(x-2\right)\left(x+2\right)\left(2x-1\right)$