Question

if x square - 1 is a factor of ax power 4 + bx cube + cx square + dx + e , show that a + c + e = 0 , b +d = 0

Answer 1

^2 and^3 is square and cube 
so lets start the question 
x^2-1=0
x^2-1^2=0
(x+1)(x-1)=0
x+1=0
x=-1
x-1=0
x=1
p(x)=ax^4+bx^3+cx^2+dx+e
p(-1)=a(-1)^4+b(-1)^3+c(-1)^2+d(-1)+e
= a-b+c-d+e-------------(i)
p(1)=a(1)^4+b(1)^3+c(1)^2+d(1)+e
a+b+c+d+e--------------(ii)
add(1)&(2)
a-b+c-d+e+a+b+c+d+e
2a+2c+2e=0
2(a+c+e)=0
a+c+e=0/2
a+c+e=0
substract1&2
a-b+c-d+e-(a+b+c+d+e)
a-b+c-d+e-a-b-c-d-e
-2b-2d=0
-2(b+d)=0
b+d=0/-2
b+d=0
hence verified

Answer 2

Answer:

refer explanation for proof a + c + e = 0 and  b +d = 0

Step-by-step explanation:

Given : $x^2-1$ is a factor of $ax^4+bx^3+cx^2+dx+e$

we have to show a + c + e = 0 , b + d = 0

Since $x^2-1$ is a factor of $p(x)=ax^4+bx^3+cx^2+dx+e$ then  $x^2-1$ must satisfy the polynomial p(x).

Put  $x^2-1=0$

We  get, $x^2=1\\\\ \rightarrow x=\sqrt{1}\\\\ \rightarrow x=\pm{1}$

Thus, at x = 1 p(x) becomes,

P(1) = 0

$p(1)=a+b+c+d+e=0$   ............(1)

also, P(-1) = 0

$p(-1)=a-b+c-d+e=0$   ............(2)

Adding (1) and (2) , we get,

a + b + c + d + e + a - b + c - d + e = 0

a + c  + e + a + c + e = 0

2(a + c+ e) = 0

⇒ a + c + e = 0

Subtract (1) and (2) , we get,

a + b + c + d + e - a + b - c + d - e = 0

b + d + b + d = 0

2( b + d) = 0

⇒ b + d = 0

Hence, proved.