Polyomials question (1 Viewer)

[kkk579]

is it because since the polynomial is now negative then dividing by x^2-5 which is positive would turn the quotient aka Q(x) negative? unless i’m mistaken

 

[Average Boreduser]

when it says p(-x)=____Q(-x)_____
i thought that you would only sub -x into ‘x^2-5’and ‘x’

Q(x) is a part of the equation though?

Q(x) is just considered apart of the polynomial- meaning that P(-x)=...Q(-x)...

P(x)+P(-x)={(x^2-5)Q(x)+(x+4)}+{(-x)^2-5)Q(x)+(-x+4)}
then we can factorise the (x^2-5)
therefore,(x^2-5)(Q(x)+Q(-x))+(x+4)+(-x+4)
try letting Q(x)=x
consider (Q(x)+Q(-x))
Q(x)+Q(-x)=x-x=0---->this applies for any equation you let Q(x) be
hence, (x^2-5)(0)+8
=8

 

[kkk579]

Q(x) is a part of the equation though?
View attachment 37364
Q(x) is just considered apart of the polynomial- meaning that P(-x)=...Q(-x)...

P(x)+P(-x)={(x^2-5)Q(x)+(x+4)}+{(-x)^2-5)Q(x)+(-x+4)}
then we can factorise the (x^2-5)
therefore,(x^2-5)(Q(x)+Q(-x))+(x+4)+(-x+4)
try letting Q(x)=x
consider (Q(x)+Q(-x))
Q(x)+Q(-x)=x-x=0---->this applies for any equation you let Q(x) be
hence, (x^2-5)(0)+8
=8

thanks, lol idk wtf i was thinking like 10 minutes ago