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Question

if a polynomial is defined as P(x) = 2x​5+ ax4+ bx3+ cx2+ dx + e such that​P(0) = 4,​P(1) = 5,​P(2) = 8,
P(3) = 13,P(4) = 20 find the value ofP(5).

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Solution

Px= 2x5+ax4+bx3+cx2+dx+eP0=0+0+0+0+0+e = 4So we get e = 4Now P1= 2+a+b+c+d+4=5or a+b+c+d=-1......2P2= 128+16a+8b+4c+2d+4=5or 16a+8b+4c+2d=-127......2Also, P3=132×35+34×a+33×b+32×c+3d+4=1381a+27b+9c+3d=-477........3P4= 20 2×45+a×44+b×43+c×42+d×4+4=20 256a+64b+16c+4d=-496......4We can see that we have 4 equation in 4 variable so we can simplify them to get the value of a,b,c,d.Try to solve it for your own practice.

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