Respuesta :

mjankiewicz
First way
n = -5n + 6n

[tex]10n^2+n-3=10n^2-5n+6n-3=5n(2n-1)+3(2n-1)=(2n-1)(5n+3)[/tex]   

Second way
[tex]f(x)=a(x-x_1)(x-x_2)[/tex] the formula

[tex]an^2+bn+c=0\\\\ 10n^2+n-3=0[/tex]

a=10  ,  b=1  ,  c=-3

[tex]\Delta=b^2-4ac=1^2-4\cdot 10\cdot (-3)=121\\\\ \sqrt\Delta=11[/tex]

[tex]n_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-1-11}{2\cdot 10}=\frac{-12}{20}=-\frac{3}{5}\\\\ n_2=\frac{-b+\sqrt\Delta}{2a}=\frac{-1+11}{2\cdot 10}=\frac{10}{20}=\frac{1}{2}[/tex]

[tex]f(n)=10[n-(-\frac{3}{5})]\cdot (n-\frac{1}{2})\\\\ f(n)=10(n+\frac{3}{5})(n-\frac{1}{2})[/tex]
the answer 1 
or

a = 10 = 5*2

[tex]f(n)=5(n+\frac{3}{5})\cdot 2(n-\frac{1}{2})\\\\f(n)=(5n+3)(2n-1)[/tex]
the answer 2

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