Question 9.4: Show that the equation x³ − 7x² + 19x − 13 = 0 has a root at......
Show that the equation x³ − 7x² + 19x − 13 = 0 has a root at x = 1 and find the other roots.
If we let P(x) = x³ − 7x² + 19x − 13, then P(1) = 1 − 7 + 19 − 13 = 0 so that x = 1 is a root. This means that x − 1 must be a factor of P(x) and so we can express P(x) in the form
\begin{aligned}P(x)=x^3-7 x^2+19 x-13 & =(x-1)\left(\alpha x^2+\beta x+\gamma\right) \\& =\alpha x^3+(\beta-\alpha) x^2+(\gamma-\beta) x-\gamma\end{aligned}
where α , β and γ are coefficients to be determined. Comparing the coefficients of x³ we find = 1. Comparing the constant coefficients we find γ = 13. Finally, comparing coefficients of x we find β = -6, and hence
P(x)=x^3-7 x^2+19 x-13=(x-1)\left(x^2-6 x+13\right)
The other two roots of P(x) = 0 are found by solving the quadratic equation x² −6x+13 = 0, that is
x=\frac{6 \pm \sqrt{36-52}}{2}=\frac{6 \pm \sqrt{-16}}{2}=3 \pm 2 \mathrm{j}
and again we note that the complex roots occur as a complex conjugate pair. This illus-trates the general result given in Section 1.4 that an nth-degree polynomial has n roots.