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I'm new to the site, and I need a bit of help from you.

How can I prove that the polynomial: $f(x)=x^3-3x^2+4x-2$ cannot be factored as a product of polynomials of degree 1 with real coefficients?

Thanks.

Asaf Karagila
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Undergrad1992
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  • Just a little note that your title is inaccurate...the polynomial CAN be factored over $\mathbb{R}$ but not in this special way as asked in the question. – fretty Dec 05 '14 at 09:08

5 Answers5

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If $f$ had $3$ real roots, then by Rolle's theorem its derivative would have $2$ real roots, but $f'(x) = 3x^2 - 6x + 4$ has no real roots because its discriminant is negative.

Dylan Yott
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By the rational root theorem, the possible rational roots are $\pm1$, and $\pm2$. Since $f(1)=0$, $(x-1)$ must be a factor of $f$. After dividing we have that

$$f(x)=x^3−3x^2+4x−2=(x-1)(x^2-2x+2).$$

Now we can take the discriminant of the quadratic factor $(x^2-2x+2)$, that being

$$b^2-4ac=(-2)^2-4\cdot 1 \cdot 2 = -4,$$

to note that $f$ has one real root and two complex roots. Hence $f$ can only be factored as the product of one linear factor and the two complex conjugate linear factors, that being

$$f(x)=(x-1)(x-(1+i))(x-(1-i)).$$

J. W. Perry
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  • Another clean way of recognizing that the quadratic factor $x^2-2x+2$ has no real roots (which is the same as computing the discriminant, but might be easier to spot) is to rewrite it as $(x^2-2x+1)+1 = (x-1)^2+1$ which clearly has no real roots. – Steven Stadnicki Dec 05 '14 at 08:23
  • Absolutely and Indeed :) – J. W. Perry Dec 05 '14 at 08:25
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First note that $(x-1)$ is a factor of $f(x)$ since $f(1)=1-3+4-2=0$. Therefore, we have $f(x)=(x-1)(Ax^2+Bx+C)$. Compare the coefficients, we can find that $$f(x)=(x-1)(x^2-2x+2).$$ However, the factor $x^2-2x+2$ cannot be factored as a product of polynomials of degree 1 with real coefficients, because it has discriminant of $(-2)^2-4(2)=-4<0$.

Paul
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Observe that$$\begin{align*} f(x)=x^3-3x^2+4x-2&=(x^3-3x^2+3x-1)+(x-1)\\\\ &=(x-1)^3+(x-1)\\\\ &=(x-1)((x-1)^2+1)\\\\ &=(x-1)(x^2-2x+2) \end{align*}$$ By the quadratic formula, the roots of $x^2-2x+2$ are complex, so $f$ cannot be factored any further into polynomials with real coefficients.

Sal
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The derivative $3x^2-6x+4$ has no real roots, so $f$ can't have two distinct real roots nor a multiple real root.

Martín-Blas Pérez Pinilla
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