Video: Roots of Cubics

Given that 𝑖 is one of the roots of the equation π‘₯Β³ βˆ’ 5π‘₯Β² + π‘₯ βˆ’ 5 = 0, find the other two roots.

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Video Transcript

Given that 𝑖 is one of the roots of the equation π‘₯ cubed minus five π‘₯ squared plus π‘₯ minus five equals zero, find the other two roots.

The factor theorem tells us that π‘₯ minus 𝑖 is one of the factors of this polynomial. And so, we could divide the left-hand side by this factor to get the quadratic π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, which we could then solve. But we can make things easier for ourselves by using the conjugate root theorem which tells us that the complex conjugate of 𝑖 must also be a root as we’re dealing with a polynomial with real coefficients. And so, π‘₯ plus 𝑖 again by the factor theorem must be a factor of this polynomial.

Multiplying the two known factors together, we get π‘₯ squared plus one. And we can distribute again. And comparing coefficients, we can see that π‘š is one and 𝑛 is negative five. We can substitute these values then, factoring our cubic as π‘₯ minus one times π‘₯ plus one times π‘₯ minus five. Remember that we’re looking for the roots of this equation. And we can just read them off from the factored form. We find that they are five, negative 𝑖, and 𝑖. For this problem, the conjugate root theorem saved us some working, but wasn’t essential.

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