Question

Write the left side of each equation as a product of linear factors, and state the solutions.
a. x^3 − 1 = 0
b. x^3 + 8 = 0
c. x^4 + 7x^2 + 10 = 0

Answer 1

Answer:

a). $x_{1}=1$, $x_{2}=1$,$x_{3}=1$

b). $x_{1}=2$, $x_{2}=2$, $x_{3}=2$

c). $x_{1} = \sqrt{-5}$, $x_{2} = \sqrt{-5}$, $x_{2} = \sqrt{-2}$,$x_{4} = \sqrt{-2}$

Step-by-step explanation:

a.

$x^{3}-1=0\\x^{3}=1\\ x=(1 )^{\frac{1}{3} }$

$x=1$

b.

$x^{3}+8=0\\x^{3}=-8\\ x=(-8 )^{\frac{1}{3} }$

$x= 2i$

c.

$x^{4} +7x^{2} +10=0\\x^{4}=u^{2} \\x^{2}=u\\ u^{2}+7u+10=0$

Using

$\frac{-b+/-\sqrt{7^{2}-4*a*c } }{2*a}$

$u=\frac{-7+/-\sqrt{7^{2}-4*10}}{2}=-\frac{7}{2} +/- \frac{\sqrt{49-40} }{2} \\ u=-\frac{7}{2} +/- \frac{3}{2} \\u_{1}= -5\\u_{2}= -2$

But u is no the factor is x so:

$x=\sqrt{u}\\ x_{1} =\sqrt{-5} =5i\\ x_{2} =\sqrt{-2} =2i$

Check:

$(\sqrt{5i})^{4} +7*(\sqrt{5i})^{2}  +10=0\\25*i^{4} +7*5*i^{2}+10=0\\ 25*1+7*5*-1+10\\25-35+10=0\\0=0$