Question

The two zeroes of the polynomial P(x)=x^3-mx^2+nx - 6 are 1 and 2.
The values of m =
and n =​

Answer 1

Given equation

$x^{3}-mx^{2} +nx-6=0$

1 and 2 are the zeroes of the polynomial.

Substituting x = 1 in the equation, we get,

$\implies (1)^{3}-m(1)^{2}+n(1) - 6 = 0\\\implies 1-m+n-6=0\\ \implies -m+n-5=0\\\implies -m+n = 5\\\implies m-n=-5 -- (1)$

Substituting x = 2 in the equation, we get,

$\implies (2)^{3} -m(2)^{2} +n(2) -6=0\\ \implies 8-4m+2n-6=0\\\implies -4m+2n-2=0\\\implies -2m+n-1=0\\\implies -2m+n = 1\\\implies 2m-n=-1--(2)$

Solving, (1) and (2), we get,

$m-n=-5\\\underline{2m-n=-1}\\\underline{\underline{-m = -4}}\\\implies m = 4$

Substituting the value of 4 in the first equation, we get,

$4-n=-5\\\implies -n=-5-4\\\implies -n=-9\\\implies n = 9$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ m \ and \ n \ are \ 4 \ and \ 9 \ respectively}}}}}}}$

                                                         

Answer 2

Step-by-step explanation:

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