Question

me podrian ayudar a resolverlo porfavor??

Answer 1

Me podrían ayudar a resolverlo por favor??

$P(x)= (x^3-5x+5) ^4 - 2(x+3)^2+8 \\ \\ \\ P(x)= (x^3-5x+5) ^2*(x^3-5x+5)^2 - (2x+3)^2+8 \\ \\--------------------------\\ Calculos \ auxiliares \to (x^3-5x+5)^2= (x^3-5x+5)(x^3-5x+5) \\ \\ (x^3-5x+5)^2= (x^3)^2+(x^3*(-5x)) + x^3*5 -5x*x^3 \\ -5x(-5x)-5x*5+5x^3-5*5x+5*5 \\ \\ (x^3-5x+5)^2= x^6-5x^4+5x^3-5x^4+25x^2-25x+5x^3-25x+25 \\ \\ (x^3-5x+5)^2= x^6-10x^4+10x^3+25x^2-50x+25 \\--------------------------$


$P(x)= (x^3-5x+5) ^4 - 2(x+3)^2+8 \\ \\ P(x)= (x^3-5x+5) ^2*(x^3-5x+5)^2 - 2(x+3)^2+8 \\ \\ P(x) = (x^6-10x^4+10x^3+25x^2-50x+25)*(x^6-10x^4+10x^3 \\ +25x^2-50x+25) - 2(x^2+6x+9) + 8= \\ \\ \\ P(x) = x^1^2-20x^1^0+20x^9 +150x^8-300x^7- 350x^6 +1500x^5 \\ -875x^4-2000x^3+3750x^2-2500x+625 - 2x^2-12x-18 +8\\ \\ \\ P(x) = x^1^2-20x^1^0+20x^9 +150x^8-300x^7- 350x^6 +1500x^5 \\ -875x^4-2000x^3+3748x^2-2512x+615$


Espero que te sirva, salu2!!!!