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...

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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

 

 

 

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