Given a function $p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f$. Each coefficient $a, b, c, d, e$, and$ f$ is equal to either $ 1$ or $-1$. If $p(2) = 11$, what is the value of $p(3)$?
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Tags: algebra, polynomial
Bratin_Dasgupta
21.09.2021 20:15
https://www.chegg.com/homework-help/questions-and-answers/1-find-sum-integers-n-property-n-n-2021-perfect-squares-2-given-function-p-x-ax-5-bx-4-cx--q83106499
parmenides51
21.09.2021 20:37
Bratin_Dasgupta wrote: https://www.chegg.com/homework-help/questions-and-answers/1-find-sum-integers-n-property-n-n-2021-perfect-squares-2-given-function-p-x-ax-5-bx-4-cx--q83106499 I believe that there is no point in linking to a non free source, as we cannot see the answer without having registered
climbmountain
22.09.2021 01:33
$p(2)=32a+16b+8c+4d+2e+f=11$ Note that $32-16-8+4-2+1=11$ Thus, $p(x)=x^5-x^4-x^3+x^2-x+1$ $p(3)=243-81-27+9-3+1=142$
OlympusHero
22.09.2021 01:59
We have $32-16-8+4-2+1=11$, so our polynomial is $x^5-x^4-x^3+x^2-x+1$. Our answer is $3^5-3^4-3^3+3^2-3+1 = \boxed{142}$.
#YUH
kante314
22.09.2021 02:08
We have
$$32a+16b+8c+4d+2e+f=11$$After testing some we see that $$32-16-8+4-2+1=11$$Thus our polynomial is
$$x^5-x^4-x^3+x^2-x+1$$Plugging in $x=3$ gives
$$3^5-3^4-3^3+3^2-3+1=243-81-27+9-3+1=\boxed{142}$$
Arrowhead575
22.09.2021 02:41
Together we ascertain that this simplifies to $32-16-8+4-2+1=11$, so our quintic is $k^5-k^4-k^3+k^2-k+1$. Our answer is $3^5-3^4-3^3+3^2-3+1 = \boxed{142}$.
#YEUH
OofPirate
22.09.2021 06:48
Note that the function when $p(2)$ is equal to $p(2)=32a+16b+8c+4d+2e+1$. By some means, I had found that $a=1$, $b=-1$, $c=-1$, $d=1$, $e=-1$, and $f=1$. Hence, the following is the function: $p(x)=x^5-x^4-x^3+x^2-x+1$. $p(3)=243-81-27+9-3+1$. $p(3)=\boxed{142}$.