MHB Solving a Polynomial x^6 – 7x^3 + 12 by Factoring.

[RidiculousName]

Hello, I have been going through the Wisconsin Placement Exam sample test. I'm trying to figure out how to find the solution set for x⁶ – 7x³ + 12.

I have tried having u = x³ and solving for u²-7u+12, but I'm unsure what to do once I get (u - 4)(u - 3).

Would someone help me figure out how to solve this type of problem?

 

[skeeter]

Hello, I have been going through the Wisconsin Placement Exam sample test. I'm trying to figure out how to find the solution set for x⁶ – 7x³ + 12.

I have tried having u = x³ and solving for u²-7u+12, but I'm unsure what to do once I get (u - 4)(u - 3).

Would someone help me figure out how to solve this type of problem?

if $x^6-7x^3+12 = 0$ ...

$u = x^3 \implies u^2-7u+12 = 0 \implies (u-4)(u-3) = 0$

the zero product property $\implies u = 4$ or $u = 3$

$u = 4 \implies x^3 = 4 \implies x = \sqrt[3]{4}$

$u = 3 \implies x^3 = 3 \implies x = \sqrt[3]{3}$

 

[RidiculousName]

if $x^6-7x^3+12 = 0$ ...

$u = x^3 \implies u^2-7u+12 = 0 \implies (u-4)(u-3) = 0$

the zero product property $\implies u = 4$ or $u = 3$

$u = 4 \implies x^3 = 4 \implies x = \sqrt[3]{4}$

$u = 3 \implies x^3 = 3 \implies x = \sqrt[3]{3}$

Thank you!