Given:
[tex]y^4-3y^3-2y^2+10y-12=0[/tex]Required:
We need to find roots of the equation.
Explanation:
[tex]\text{ The factors of 12 are P= \textbraceleft}\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\rbrace[/tex][tex]The\text{ factors of 1 are Q=\textbraceleft}\pm1\rbrace[/tex][tex]\frac{P}{Q}=P=\text{\textbraceleft}\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\rbrace[/tex]The possible roots of the equation are
[tex]\text{\textbraceleft}\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\rbrace[/tex]Let y =1 and substite int the equation.
[tex](1)^4-3(1)^3-2(1)^2+10(1)-12=0[/tex][tex]1-3-2+10-12=0[/tex][tex]-6=0[/tex]y=1 is not a root.
Let y =-1 and substitue in the equation.
[tex](-1)^4-3(-1)^3-2(-1)^2+10(-1)-12=0[/tex][tex]1+3-2-10-12=0[/tex][tex]-20=0[/tex]y= -1 is not a root.
Let y = 2 and substitue in the equation.
[tex](2)^4-3(2)^3-2(2)^2+10(2)-12=0[/tex][tex]16-24-8+20-12=0[/tex][tex]-8=0[/tex]y=2 is not a root.
Let y = - 2 and substitue in the equation.
[tex](-2)^4-3(-2)^3-2(-2)^2+10(-2)-12=0[/tex][tex](2)^4+3(2)^3-2(2)^2-10(2)-12=0[/tex][tex]16+24-8-20-12=0[/tex][tex]0=0[/tex]y = -2 is the root of the given equation.
Use the synthetic method to find the roots.
The given equation can be written as follows.
[tex](y+2)(y^3-5y^2+8y-6)=0[/tex]Consider the equation to find the remaining roots.
[tex]y^3-5y^2+8y-6=0[/tex]Set y = 2 ans substitute in the equation.
[tex](2)^3-5(2)^2+8(2)-6=0[/tex][tex]8-20+16-6=0[/tex][tex]-2=0[/tex]y =2 is not the second root.
Set y = -2 and substitute in the equation.
[tex](-2)^3-5(-2)^2+8(-2)-6=0[/tex][tex]-(2)^3-5(2)^2-8(2)-6=0[/tex][tex]-50=0[/tex]y = -2 is not the second root.
Set y = 3 and substitute in the equation.
[tex](3)^3-5(3)^2+8(3)-6=0[/tex][tex]27-45+24-6=0[/tex][tex]0=0[/tex]y = 3 is the second root.
Use the synthetic method.
The given equation can be written as follows.
[tex](y+2)(y-3)(y^2-2y+2)=0[/tex]Consider the equation to find the remaining roots.
[tex]y^2-2y+2=0[/tex]Which is of the from
[tex]ax^2+bx+c=0[/tex]where a =1, b=-2 and c=2.
Cnmdidr the qudaratic formuola.
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Substitute a =1, b=-2 and c=2 in the formula.
[tex]y=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(1)}}{2(1)}[/tex][tex]y=\frac{2\pm\sqrt{4-8}}{2}[/tex][tex]y=\frac{2\pm\sqrt{i^22^2}}{2}[/tex][tex]y=\frac{2\pm2i}{2}[/tex][tex]y=1+i,1-i[/tex]Final answer:
The roots are -2, -3,1+i and 1-i.
