Video: Roots of Quartics

Given that 2 + π‘–βˆš3 is a root of π‘₯⁴ βˆ’ 12π‘₯Β³ + 55π‘₯Β² βˆ’ 120π‘₯ = 112 = 0, find all the roots.

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Video Transcript

Given that two plus 𝑖 root three is a root of π‘₯ to the power of four minus 12π‘₯ cubed plus 55π‘₯ squared minus 120π‘₯ plus 112 equals zero, find all the roots.

We have a complex root of a polynomial with real coefficients. And so, by the conjugate root theorem, its complex conjugate two minus 𝑖 root three is also a root of this equation. We found two roots then. But how do we find the other two? The factor theorem gives us two linear factors of our quartic from these two roots. What can we say about the remaining factor? Well, it must be quadratic so that distributing our right-hand side gives a quartic as on the left. And the roots of this unknown quadratic factor are the remaining roots of our quartic equation. Let’s set about finding this quadratic factor then.

We multiply the two linear factors together to get a known quadratic factor. There are a couple of identities which reduce the amount of calculation required. Now, we can just divide our quartic by this known quadratic factor to find the unknown quadratic factor. Alternatively, we can distribute on the right-hand side multiplying every term in the first set of parentheses by every term in the second set and then collecting like terms. We can then compare coefficients.

For example, the coefficient of π‘₯ to the four tells us that π‘Ž is one. We make the substitution and then compare the coefficients of π‘₯ cubed, finding that 𝑏 minus four is negative 12. And so, 𝑏 is negative eight. Again, making the substitution, we then compare the coefficients of π‘₯ squared, finding that 𝑐 is 16. And making the substitution really does make the right-hand side equal to the left. We found the unknown quadratic factor then. It’s one π‘₯ squared minus eight π‘₯ plus 16. And we can factor this as π‘₯ minus four squared. It has repeated root of four. And so, the original quartic must do too.

Having factored our quartic, we read off the roots. The roots are four, repeated root with multiplicity two; two plus 𝑖 root three, the complex root that we were given; and its complex conjugate two minus 𝑖 root three.

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