If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$ , then show that the matrix A is a root of the polynomial $f (x) = x^{3} - 6 x^{2} + 7 x + 2$ .

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ = 1 (6 - 0) - 0 + 2 (0 - 4) = 6 - 8 = - 2 f (x) = x^{3} - 6 x^{2} + 7 x + 2 = (x - 2) (x^{2} - 4 x - 1)$
Hence $A$ is not root of polynomial $f (x)$

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Similar questions

A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦

x^{3} - 6 x^{2} + 7 x + k = 0

k

f (x) = x^{3} + 6 x^{2} + 11 x + 6

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x)

f(x) = $x^{3} - 6 x^{2}$ + 11x - 6, g(x) = $x^{2}$ -3x + 2

α

β

f (x) = 6 x^{2} - 3 - 7 x

(α + 1) (β + 1)

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