Question

If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0

Answer 1

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

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

$p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\ p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\ p(1)=a+b+c+d+e \\ p(-1)=a-b+c-d+e \\ \\ p(1)=0 \\ p(-1)=0 \\ \\ \hbox{add both equations:} \\ a+b+c+d+e=0 \\ \underline{a-b+c-d+e=0} \\ 2a+2c+2e=0 \\ 2(a+c+e)=0 \\ a+c+e=0 \\ \\ \hbox{substitute 0 for a+c+e in the first equation:} \\ a+b+c+d+e=0 \\ (a+c+e)+b+d=0 \\ 0+b+d=0 \\ b+d=0 \\ \\ \boxed{a+c+e=b+d=0} \\ \hbox{proved } \checkmark$