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Algebra 2, Solving Polynomial Equations
Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18.
 one year ago
 one year ago
Algebra 2, Solving Polynomial Equations Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18.
 one year ago
 one year ago

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theslytherinhelperBest ResponseYou've already chosen the best response.0
I've gotten most of the work down, if you'd like to see it. I'm stuck on the last step :(
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
youre not doing ferrari are you?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
x4 – x3 + 7x2 – 9x – 18=0 to get rid of the x3 term, let x=(y+1/4) (y+1/4)^4 – (y+1/4)^3 + (y+1/4)^2 – 9(y+1/4) – 18=0 \[y^4+\frac58y^2\frac{69}8y\frac{5171}{256}=0\] put the linear parts on the other side \[y^4+\frac58y^2=\frac{69}8y+\frac{5171}{256}\] complete the square on the left by adding 5y^2/8+25/64 to each side \[(y^2+\frac58)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}\] this is as far as we can go without introducing another variable, like z adding a z into the left side gives us more room to play with, but we have to adjust the right side by the same amount: (a+b+c)^2 = a^2+2ab+b^2+(c^2+2c(a+b)) so lets input the z on the left, and add z^2 + 2z(y^2+5/8) to the right side \[(y^2+\frac58+z)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}+z^2 + 2z(y^2+\frac58)\] lets clean up the right side to see it in its quadraitc form for y \[(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256}) \] solving the right side for yusing the quadratic formula we get: \[\Large y=\frac{\frac{69}{8}\pm\sqrt{(\frac{69}{8})^24(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})}}{2(2z+\frac58)}\] if we can get the discriminant to equal zero, for some reason, the method will give us roots ... so \[(\frac{69}{8})^24(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})=0\] which simplifies (lol) to:\[8z^3+\frac{25}{2}z^2+\frac{5371}{32}z\frac{11733}{512}=0\]which is just Cardanos method now.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
Cardanos method is to reduse it by removing the z^2; let z=(u25/48) 8(u25/48)^3+25(u25/48)^2/2+5371(u25/48)/3211733/512 which reduces to:\[u^3+\frac{484}{24}u=\frac{2918}{216}\] hence: \[u=\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}+\frac{(2918/216)}{2}}\\~~~~~\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}\frac{(2918/216)}{2}}\] \[u=...\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
lol, im pretty sure i might have mistyped a number or two, but if youve never heard of ferrari then im sure theres a simpler way they want you to look at this
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
what is your last step?
 one year ago

theslytherinhelperBest ResponseYou've already chosen the best response.0
Thanks sooo much! The last step is simplifying the equation. ^_^
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
im not sure of the method you used, so ill need to know a little more about how you approached this
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
my u wasnt all the way simplified \[u^3+\frac{121}{6}u=\frac{1459}{108}\] might make life on my method here a smidge simpler \[u = \sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}=abt~~0.6559\] \[z=u\frac{25}{48}=\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}\frac{25}{48}\] plugging this into our (y^2...)^2 = y^2+... setup with the zs gives us \[(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256})\] \[(y^2+\frac58+(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}\frac{25}{48}))^2\\=(2(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}\frac{25}{48})+\frac58)y^2+\frac{69}8y\\+((\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}\frac{25}{48})^2\\+\frac{10}{8}(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}\sqrt[3]{\frac{\sqrt{22361}}{8}\frac{1459}{216}}\frac{25}{48})+\frac{5271}{256}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
im sooo glad for the wolfram :)
 one year ago
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