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
