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

AI Thread Summary
To solve the polynomial x^6 – 7x^3 + 12, the substitution u = x^3 simplifies the equation to u^2 - 7u + 12. Factoring gives (u - 4)(u - 3) = 0, leading to u = 4 or u = 3. This results in x^3 = 4, giving x = ∛4, and x^3 = 3, giving x = ∛3. The complete solution set for the original polynomial includes the cube roots of 4 and 3.
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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?
 
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RidiculousName said:
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 $u = 3$

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

$u = 3 \implies x^3 = 3 \implies x = \sqrt[3]{3}$
 
skeeter said:
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!
 
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