Show that x + a is a factor of the polynomial P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5. Plz help I can't fin anything online to help with this. All the examples given are with x as the only variable.
3 answers
nvm i solved it
This is the basic concept of the Factor Theorem:
If for some function f(x), x-a is a factor, then f(a) = 0
so we are given that x+a is a factor, so let's look at f(-a)
P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5 , now replace all x's with -a
P(-a) = (-a + a)^5 + (-a + c)^5 + (a - c)^5
= 0 + [(-a)^5 + 5(-a)^4 c + 10(-a)^3 c^2 + 10(-a)^2 c^3 + 5(-a) c^4 + c^5 ]
+ (a)^5 + 5(a)^4 (-c) + 10(a)^3 (-c)^2 + 10(a)^2 (-c)^3 + 5(a) (-c)^4 + (-c)^5
= 0 - a^5 + 5a^4 c - 10a^3 c^2 + 10a^2 c^3 - 5ac^4 + c^5 + a^5 - 5a^4 c + 10a^3 c^2 - 10a^2 c^3 + 5a c^4 - c^5
= 0
therefore x+a is a factor of P(x)
If for some function f(x), x-a is a factor, then f(a) = 0
so we are given that x+a is a factor, so let's look at f(-a)
P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5 , now replace all x's with -a
P(-a) = (-a + a)^5 + (-a + c)^5 + (a - c)^5
= 0 + [(-a)^5 + 5(-a)^4 c + 10(-a)^3 c^2 + 10(-a)^2 c^3 + 5(-a) c^4 + c^5 ]
+ (a)^5 + 5(a)^4 (-c) + 10(a)^3 (-c)^2 + 10(a)^2 (-c)^3 + 5(a) (-c)^4 + (-c)^5
= 0 - a^5 + 5a^4 c - 10a^3 c^2 + 10a^2 c^3 - 5ac^4 + c^5 + a^5 - 5a^4 c + 10a^3 c^2 - 10a^2 c^3 + 5a c^4 - c^5
= 0
therefore x+a is a factor of P(x)
P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5
P(-a) = (-a+a)^5 + (-a+c)^5 + (a-c)^5
= 0 - (a-c)^5 + (a-c)^5
= 0
so (x+a) is a factor of P(x)
P(-a) = (-a+a)^5 + (-a+c)^5 + (a-c)^5
= 0 - (a-c)^5 + (a-c)^5
= 0
so (x+a) is a factor of P(x)