Question

25а2+ 10а+ 1 : 5а2- 9а- 2

Answer 1

$\frac{25a^2+10a+1}{5a^2-9a-2}=?\\\\25a^2+10a+1=(5a)^2+2*5a*1+1^2=(5a+1)^2\\\\5a^2-9a-2=5(a-2)(a+0,2)\\D=(-9)^2-4*5*(-2)=81+40=121=11^2\\a_1= \frac{9+11}{2*5}= \frac{20}{10}=2\\a_2= \frac{9-11}{2*5}= \frac{-2}{10}=-0,2\\\\ \frac{25a^2+10a+1}{5a^2-9a-2}= \frac{(5a+1)^2}{5(a-2)(a+0,2)}= \frac{25(a+0,2)^2}{5(a-2)(a+0,2)}= \frac{5(a+0,2)}{a-2}$