Question

If 1 and -1 are zeros of the polynomial Lx4 + Mx3 +Nx2 +Rx + p . Show that L+N+P=M+R=0

Answer 1

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GIVEN:

• Lx⁴ + Mx³ + Nx² + Rx + P

• x = 1 and x = - 1

$\rule{200}{2}$

TO PROVE:

• L + N + P = M + R = 0

$\rule{200}{2}$

PROOF:

Substitute x = 1 in Lx⁴ + Mx³ + Nx² + Rx + P

L(1)⁴ + M(1)³ + N(1)² + R(1) + P = 0

L + M + N + R + P = 0

$\rule{200}{1}$

Substitute x = - 1 in Lx⁴ + Mx³ + Nx² + Rx + P

L(- 1)⁴ + M(- 1)³ + N(- 1)² + R(- 1) + P = 0

L - M + N - R + P = 0

L + N + P = M + R

$\rule{200}{1}$

Substitute L + N + P = M + R in L + M + N + R + P = 0

M + R + M + R = 0

2M + 2R = 0

2(M + R) = 0

M + R = 0

$\rule{200}{1}$

Substitute M + R = 0 in L + N + P = M + R

L + N + P = 0

So L + N + P = 0 = M + R

HENCE PROVED.

$\rule{200}{2}$

NOTE:

Here Lx⁴ + Mx³ + Nx² + Rx + P is equated to zero because if x = - 1 (or) + 1 is a root of the equation, when substituted x = - 1 (or) + 1 the result should equal to zero.

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Answer 2

${\red{\underline{\underline{\bold{Given:-}}}}}$

• A quadratic polynomial Lx⁴ + Mx³ + Nx²+ Rx+P

• The zeroes of the polynomial are 1 and -1

${\blue{\underline{\underline{\bold{To\:prove:-}}}}}$

• L + N + P = M + R

${\green{\underline{\underline{\bold{Proof:-}}}}}$

Given 1 and -1 are the zeroes of the polynomial

So:-

Substitute x = 1 in Lx⁴ + Mx³ + Nx²+ Rx+P

➪ L(1)⁴ + M(1)³ + N(x)² + R(x) + P

➪ L + M + N + R + P = 0..........(1)

Substitute x = -1 in Lx⁴ + Mx³ + Nx²+ Rx+P

➪L(-1)⁴ + M(-1)³ + N(-x)² + R(-x) + P

➪ L - M + N - R + P = 0

➪ L + N + P = M + R.......... (2)

Substitute equation (2) in (1)

➪L + N + P +M + R = 0

➪ M + R + M + R = 0

➪ 2(M + R) = 0

➪M + R = 0

From equation (2)

L + N + P = M + R = 0

Hence Proved