Question

Find the value of p(-2) for p(y)=y^4-2y^3+y-7

Answer 1

Hello!!

$p(y)=y{}^{4}-2y{}^{3}+y-7$

put y=(-2)

$p(-2)=(-2){}^{4}-2(-2){}^{3}+(-2)-7$

$p(-2)=16+16-2-7$

$p(-2)=32-9$

$p(-2)=23$

Hope it's help uh

Answer 2

Answer:

$\sf p(y)= y{}^{4}-2y{}^{3}+y-7$

$\sf put y=(-2)$

$\sf p(-2)=(-2){}^{4}-2(-2){}^{3}+(-2)-7$

$\sf p(-2)=16+16-2-7$

$\sf p(-2)=32-9$

$\sf p(-2)=23$