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If a, b, c d, and e are constants such that every x > 0 satisfies

\((5x^4 -8x^3 +2x^2 +4x+7)/(x+2)^4 = a+b/(x+2) +c/(x+2)^2 +d/(x+2)^3 +e/(x+2)^4\)
then what is the value of a+b+c+d+e?

 Jul 20, 2020
 #1
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+1

a + b + c + d + e = 1 + (-4) + 8 + 7 + (-3) = 9.

 Jul 21, 2020
 #2
avatar+110727 
0

(5x^4 -8x^3 +2x^2 +4x+7)/(x+2)^4 = a+b/(x+2) +c/(x+2)^2 +d/(x+2)^3 +e/(x+2)^4

 

\(\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\ \\~\\ \frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\ a=5\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ -8=40+b\\ b=-48\\~\\ ( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ \)

 

Just keep going till you have them all.

There is probably a better way though.

 

Coding:

\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\
\\~\\
\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\
a=5\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\
-8=40+b\\
b=-48\\~\\
( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\

 Jul 21, 2020
edited by Melody  Jul 21, 2020
 #4
avatar
+1

As with Melody above to,

\(\displaystyle 5x^{4}-8x^{3}+2x^{2}+4x+7\equiv a(x+2)^{4}+b(x+2)^{3}+c(x+2)^{2}+d(x+2)+e,\)

but now substitute x = -1.

Guest Jul 22, 2020
 #5
avatar+146 
+1

Thank you, all three of you, for helping me with this problem.

Firebolt  Jul 22, 2020

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