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I am stuck with the following problem:

Solve the equations by Ferrari's method: $2x^4+6x^3-3x^2+2=0$

My try: $$2x^4+6x^3-3x^2+2=0 \implies (2x^2+3x+\lambda)^2=(15+4 \lambda)x^2+(6\lambda)x+(\lambda^2-4)... \tag{1}$$

I skipped some steps.

Right hand side of $(1)$ will be a perfect square if $$(6\lambda)^2-4(\lambda^2-4)(15+4\lambda)=0 \implies 2\lambda^3+3\lambda^2-8\lambda-30=0$$.

Now I am stuck . I can't factorize the last step. Can someone help?

Thanks in advance for your time.

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By rational root test we can find

$$2\lambda^3+3\lambda^2-8\lambda-30=(2 \lambda - 5) (\lambda^2 + 4 \lambda + 6) $$

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Notice that the leading coefficient is $2$ so we must have either $$(2\lambda+\text{?})(\lambda^2+\text{?}\lambda+\text{?})\quad\text{or}\quad(\lambda+\text{?})(2\lambda^2+\text{?}\lambda+\text{?})$$Note also that $30=1\cdot30=2\cdot15=3\cdot10=5\cdot6$ so eliminating the possibilities gives $$2\lambda^3+3\lambda^2-8\lambda-30=(2\lambda-5)(\lambda^2+4\lambda+6)$$

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  • $\begingroup$ Thanks a lot. Got it.. $\endgroup$ – learner Mar 9 '18 at 16:29
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Hint:

$2\lambda^3+3\lambda^2-8\lambda-30=0 \Leftrightarrow 2\lambda^3-5\lambda^2+8\lambda^2-20\lambda+12\lambda-30=0$

$\Leftrightarrow (2\lambda-5)(\lambda^2+4\lambda+6)=0$

You only need to minimize $\lambda^2+4\lambda+6$ to finish this equation.

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No need to use Ferrari, because we have $$ 2x^4+6x^3-3x^2+2=(2x^2 - 2x + 1)(x^2 + 4x + 2).$$

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