p1=2x2+x+3
p2=5x2−x+5
p3=3x2+5x+2
x2+x+1=a1(2x2+x+3)+b1(5x2−x+5)+c1(2+5x+3x2)
x2(2a1+5b1+3c1)+x(a1−b1+5c1)+(3a1+5b1+2c1)=x2+x+1
2a1+5b1+3c1=1;a1−b1+5c1=1;3a1+5b1+2c1=1
(3a1+5b1+2c1)−(3a1+4b1+8c1)=−1
b1=6c1−1 ---(1)
2(b1−5c1+1)+5b1+3c1=1
7b1−6c1+1=0 ---(2)
Solving (1) and (2) we get
c1=6/35
Substituting this in equation(1) we get
b1=1/35;a1=6/35
So,
x2+x+1=(6/35)(2x2+x+3)+(1/35)(5x2−x+5)+(6/35)(2+5x+3x2)
Comments