"p_1=2x^2+x+3"
"p_2=5x^2-x+5"
"p_3=3x^2+5x+2"
"x^2+x+1=a_1(2x^2+x+3)+b_1(5x^2-x+5)+c_1(2+5x+3x^2)"
"x^2(2a_1+5b_1+3c_1)+x(a_1-b_1+5c_1)+(3a_1+5b_1+2c_1)=x^2+x+1"
"2a_1+5b_1+3c_1=1;a_1-b_1+5c_1=1;3a_1+5b_1+2c_1=1"
"(3a_1+5b_1+2c_1)-(3a_1+4b_1+8c_1)=-1"
"b_1=6c_1-1" ---(1)
"2(b_1-5c_1+1)+5b_1+3c_1=1"
"7b_1-6c_1+1=0" ---(2)
Solving (1) and (2) we get
"c_1=6\/35"
Substituting this in equation(1) we get
"b_1=1\/35;a_1=6\/35"
So,
"x^2+x+1=(6\/35)(2x^2+x+3)+(1\/35)(5x^2-x+5)+(6\/35)(2+5x+3x^2)"
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