Question

Necesito resolver esto yaaaaa

Answer 1

Respuesta:

Explicación paso a paso:

a) $x^3-8x^2+19x-12=0$

$factorizar \\x^3-8x^2+19x-12=0$

$\left(x-1\right)\left(x-3\right)\left(x-4\right)=0$

$x-1=0\quad \mathrm{,}\quad \:x-3=0\quad \mathrm{,}\quad \:x-4=0$

$x=1 x=3 x=4 raices de la funcion$

b)

$x^3-3x^2-34x-48=0$

$\left(x+2\right)\left(x+3\right)\left(x-8\right)=0$

$x+2=0\quad \mathrm{o}\quad \:x+3=0\quad \mathrm{o}\quad \:x-8=0$

x= -2  o x= -3  o  x=8

c)

$x^3+x^2-26x+24=0\\factorizar \\x^3+x^2-26x+24=0\\\left(x-1\right)\left(x-4\right)\left(x+6\right)=0\\x-1=0\quad \mathrm{,}\quad \:x-4=0\quad \mathrm{,}\quad \:x+6=0$

x= 1  o   x= 4  x= -6

d)

$x^3-x^2-16x-20=0\\\left(x+2\right)^2\left(x-5\right)=0\\x+2=0\quad \mathrm{or}\quad \:x-5=0\\x=-2,\:x=5$

e)

) $x^3-8x^2+19x-12=0\\\left(x-1\right)\left(x-3\right)\left(x-4\right)=0\\x-1=0\quad \mathrm{o}\quad \:x-3=0\quad \mathrm{o}\quad \:x-4=0\\x=1,\:x=3,\:x=4$

f)

$x^3-6x^2+11x-6=0\\\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\x-1=0\quad \mathrm{o}\quad \:x-2=0\quad \mathrm{o}\quad \:x-3=0\\\\x=1,\:x=2,\:x=3$

g)

$x^3-6x^2+3x+10=0\\\left(x+1\right)\left(x-2\right)\left(x-5\right)=0\\x+1=0\quad \mathrm{o}\quad \:x-2=0\quad \mathrm{o}\quad \:x-5=0\\x=-1,\:x=2,\:x=5$

h) es igual a la g)

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