# If (x2-1) is factor of ax4+bx3+cx2+dx+e, show that a+b+c+d+e=0 and a+c+e=b+d

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I suppose you mean:

x^2-1 is a factor for

ax^4+bx^3+cx^2+d^x+e

show that:

1) a+b+c+d+e=0

X^2-1 is a factor means that x^2-1 zero are the function's zeros too.

x^2-1=0

==> x1= 1 x2=-1

Now by substituting in the function:

x=1 ==> a(1)+b(1)+c(1)+d(1)+e=0

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

2) a+c+e = b+d

To prove that we need to substitute with x=-1 (the other root)

==> a(-1)^4+b(-1)^3+c(-1)^2+d(-1)+e=0

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

==> a+c+e = b+d

Supposing that you mean that (x^2 - 1) is a factor of ax^4+bx^3+cx^2+dx+e, that means also that the roots of the factor x^2-1 are the roots of the polynomial ax^4+bx^3+cx^2+dx+e, too.

So, let's find the roots of x^2-1=0

We notice that is a difference of squares:

a^2 - b^2 = (a-b)(a+b)

x^2-1 = (x-1)(x+1) = 0

Now we'll put each factor zero:

x-1 = 0

x=1

x+1 = 0

x = -1

So, as we've said before, the roots of x^2-1 are the roots of the given polynomial, meaning that:

f(-1) = 0, where f(x) = ax^4+bx^3+cx^2+dx+e

f(-1) = a-b+c-d+e = 0 (1)

f(1) = a+b+c+d+e = 0 (2)

From the relation (1), we'll get:

a+c+e = b+d

a+b+c+d+e = 0

To prove if x^2-1 is factor of f(x) = ax^4+bx^3+cx^2+dx+e.....(1) then

a+b+c+d = 0 and a+c+e = b+d.

Solution:

If x^2-1 is a factor, then (x-1) and (x+1) are also the factors of the given expression at (1)

So for x=1 , f(1) = 0 and f(-1) = 0. This gives: For x=1,a*1^4+b*1^3 + c*1^2+d*1+e = 0. Or

a+b+c+d+e = 1

Eor x = -1, f(-1) = 0.This gives: a*(-1)^4+b*(-1)^3+c*(-1)^2+ d*(-1)+e = 0. Or a-b+c-d+e = 0. Or

1+c+e+ = b+d.