Question

if 1 and - 1 are zeroes of polynomial lx4+mx3+nx2+rx+p, show that l+n+p=m+r=0

Answer 1

1 is zero of polynomial.
substitute 1 in place of x and equate to zero
L(1)^4+M(1)^3+N(1)^2+R(1)+P=0
L+M+N+R+P=0----------(1)
-1 Is zero of polynomial
sub -1 in place of x
L-M+N-R+P=0
L+N+P=M+R----(2)
From 1&2 we get L+N+P=M+R=0

Answer 2

Given:

1 and - 1 are zeroes of polynomial

$l{x}^{4} + m{x}^{3} + n{x}^{2} + rx + p$

To show:

l + n + p = m + r = 0.

Solution:

First of all, we should know that if a polynomial p(x) has zeroes a and b, then it will satisfy the polynomial p(x). This means,

p(a) = 0 and p(b) = 0

So, as given, we have,

1 and - 1 are zeroes of polynomial

${lx}^{4} + {mx}^{3} + {nx}^{2} + rx + p$ (i)

On putting x = 1 in (i), we have,

$l + m + n + r + p = 0$ (ii)

Similarly, on putting x = -1 in (i), we have,

$l{(-1)}^{4} + m{(-1)}^{3} + n{(-1)}^{2} + r(-1) + p = 0$

$l - m + n - r + p = 0$ (iii)

Now,

after adding (ii) and (iii), we have,

$2l + 2n + 2p = 0$

$l + n + p = 0$ (iv)

Similarly, after subtracting (ii) from (iii), we have,

$2m + 2r = 0$

$m + r = 0$ (v)

Hence, from (iv) and (v),

l + n + p = m + r = 0.