Let us express the following polynomial as a linear combination of the polynomials.
P=aP1+bP2+cP3
It follows that
t2+4t−3=a(t2+2t+5)+b(2t2−3t)+c(t+3),
and hence
t2+4t−3=(a+2b)t2+(2a−3b+c)t+(5a+3c)
We get the following system:
⎩⎨⎧a+2b=12a−3b+c=45a+3c=−3
⎩⎨⎧a=1−2b2(1−2b)−3b+c=45(1−2b)+3c=−3
⎩⎨⎧a=1−2b−7b+c=2−10b+3c=−8
⎩⎨⎧a=1−2bc=2+7b−10b+3(2+7b)=−8
⎩⎨⎧a=1−2bc=2+7b11b=−14
⎩⎨⎧a=1139c=−1176b=−1114
We conclude that
P=1139P1−1114P2−1176P3
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