X³-6x²+11x-6:3 find the zeros of Polynomial
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Sia ? 6 years, 6 months ago
The given polynomial f(x)={tex}\text{x}^3-\text{6x}^2+\text{11x-6}{/tex}
Since 3 is a zero of p(x), so (x - 3) is a factor of f(x).
On dividing f(x) by (x - 3), we get
{tex}\therefore{/tex} f(x) = (x2 - 3x + 2)(x - 3)
= ( x2 - 2x - x + 2)( x - 3)
= [x(x - 2) -1(x - 2)](x - 3)
= (x - 1)(x - 2)(x - 3)
Now f(x)=0 if x - 1 = 0 or x - 2 = 0 or x - 3 = 0
{tex}\Rightarrow{/tex} x = 1 or x = 2 or x = 3
{tex}\mathrm{Hence}\;\mathrm{the}\;\mathrm{remainig}\;\mathrm{roots}\;\;\mathrm{of}\;\mathrm f(\mathrm x)\;\mathrm{are}\;1\;\mathrm{and}\;2\;{/tex}
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