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.
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)
Well, let's see if we can help you out with a little laugh!
To show that x + a is a factor of P(x), we need to demonstrate that P(x) is divisible by x + a.
Now, let's start by using the Binomial Theorem. According to it, (x + y)^n can be expanded as the sum of the terms:
C(n, 0) * x^n * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ... + C(n, n-1) * x^1 * y^(n-1) + C(n, n) * x^0 * y^n
Where C(n, r) denotes the binomial coefficient, which can be calculated as C(n, r) = n! / (r! * (n-r)!).
Now, let's observe the polynomial P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5. We'll focus on the first term, (x + a)^5:
(x + a)^5 = C(5, 0) * x^5 * a^0 + C(5, 1) * x^4 * a^1 + C(5, 2) * x^3 * a^2 + C(5, 3) * x^2 * a^3 + C(5, 4) * x^1 * a^4 + C(5, 5) * x^0 * a^5
Now, let's look at each term individually:
C(5, 0) * x^5 * a^0 = x^5
C(5, 1) * x^4 * a^1 = 5 * x^4 * a
C(5, 2) * x^3 * a^2 = 10 * x^3 * a^2
C(5, 3) * x^2 * a^3 = 10 * x^2 * a^3
C(5, 4) * x^1 * a^4 = 5 * x * a^4
C(5, 5) * x^0 * a^5 = a^5
Now, if we sum all these terms, we get:
(x + a)^5 = x^5 + 5 * x^4 * a + 10 * x^3 * a^2 + 10 * x^2 * a^3 + 5 * x * a^4 + a^5
Now, let's focus on P(x) again:
P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5
We can substitute (x + a)^5 with the expanded form we obtained:
P(x) = x^5 + 5 * x^4 * a + 10 * x^3 * a^2 + 10 * x^2 * a^3 + 5 * x * a^4 + a^5 + (x + c)^5 + (a – c)^5
Now, let's combine like terms:
P(x) = x^5 + (x + c)^5 + (a – c)^5 + 5 * x^4 * a + 10 * x^3 * a^2 + 10 * x^2 * a^3 + 5 * x * a^4 + a^5
Now, if we focus on the term (x + c)^5, we can apply the same logic as before:
(x + c)^5 = x^5 + 5 * x^4 * c + 10 * x^3 * c^2 + 10 * x^2 * c^3 + 5 * x * c^4 + c^5
Let's substitute this back into P(x):
P(x) = x^5 + (x^5 + 5 * x^4 * c + 10 * x^3 * c^2 + 10 * x^2 * c^3 + 5 * x * c^4 + c^5) + (a – c)^5 + 5 * x^4 * a + 10 * x^3 * a^2 + 10 * x^2 * a^3 + 5 * x * a^4 + a^5
Now, let's combine like terms again:
P(x) = 2 * x^5 + 5 * x^4 * (a + c) + 10 * x^3 * (a^2 + c^2) + 10 * x^2 * (a^3 + c^3) + 5 * x * (a^4 + c^4) + (a^5 + c^5) + (a – c)^5
As you can see, we have the term (a + c) multiplied by x^4. Therefore, x + a is a factor of P(x)!
I hope this explanation brought a smile to your face, but more importantly, I hope it clarified the process for you!
To show that (x + a) is a factor of the polynomial P(x) = (x + a)^5 + (x + c)^5 + (a – c)^5, we need to verify if P(-a) equals zero.
Let's substitute x = -a into the polynomial and simplify:
P(-a) = (-a + a)^5 + (-a + c)^5 + (a - c)^5
= 0^5 + (-a + c)^5 + (a - c)^5
= (c - a)^5 + (a - c)^5
Now, observe that (c - a) + (a - c) equals zero, so their fifth powers will also sum up to zero. Therefore:
P(-a) = (c - a)^5 + (a - c)^5
= 0
Since P(-a) equals zero, we can conclude that (x + a) is a factor of the polynomial 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)