4 The cubic equation $x^3-2x^2+3x+4=0$ has roots $\alpha,\beta,\gamma.$

(I) Write down the values of $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, and $\alpha\beta\gamma$.

The cubic equation $x^3+px^2+10x+q=0$, where $p$ and $q$ are constants, has roots $\alpha+1$, $\beta+1$, $\gamma+1$.

(II) Find the value of $p$. $\color{gray}{-(\overbrace{\alpha+\beta+\gamma}+3)}$

(III) Find the value of $q$.

In 4.II, I know you need to use the sum of the roots given so you'd get what I have written (in gray), but how can you use $-b/a$ if you have "p" as your b term?

  • 1
    $\begingroup$ Are you actually in an exam? $\endgroup$ – user117644 Mar 30 '15 at 11:59
  • $\begingroup$ Nope haha, It's a sheet that was given to me ages ago. $\endgroup$ – JayVB Mar 30 '15 at 12:17
  • $\begingroup$ What's the relevance of $-b/a$? FWIW, the answer for 4.(ii) marked on the sheet is correct: since it's a monic polynomial (the leading coefficient is 1), $p$ is simply the sum of the roots. $\endgroup$ – PM 2Ring Mar 30 '15 at 12:18
  • $\begingroup$ I wrote that but I need to use it to find p which apparently equals -5 $\endgroup$ – JayVB Mar 30 '15 at 12:30
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    $\begingroup$ $p=-((\alpha+1)+(\beta+1)+(\gamma+1))=-(\alpha+\beta+\gamma)-3$, but $\alpha+\beta+\gamma=2$ because we are told $\alpha,\beta,\gamma$ are the roots of $x^3-2x^2+3x+4$. $\endgroup$ – Nathanson Mar 30 '15 at 13:15

If $f(x)$ of degree $3$ has roots $α,β,γ$, then $g(x)=f(x-1)$ has roots $α+1,β+1,γ+1$. Note that $$ p(x)=c_3(x-α)(x-β)(x-γ) $$ relating coefficients and elementary symmetric polynomials in the roots.

See also the root laws of Viete.


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