Question

if 1 and -1 are zeroes of the polynomial Lx⁴+Mx³+Nx²+Rx+P , show that L+N+P= M+R= 0.


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Answer 1

♥️ᎻᏆ FᎡᏆᎬNᎠ♥️

Lx^4 + Mx^3 + Nx^2 + Rx + p


P( x ) = Lx^4 + Mx^3 + Nx^2 + Rx + P

P( -1 ) = L( -1 )^4 + M( -1 )^3 + N( -1 )^2 + R( -1 ) +P = 0 ( Given -1 is zero)

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

When X= 1

P( x) = Lx^4 + Mx^3 +Nx^2 + Rx + Ꮲ

P( 1 ) = L(1 )^4 + M(1)^3 +N(1)^2 + R(1 ) + P = 0 ( give that 1 is zero)

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

ᎪᎠᎠᏆNᏩ 1 ᎪNᎠ 2

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

=>2L + 2N + 2P = 0

=> 2( L + N + P ) = 0

=> L + N + P = 0/2

=> L + N + P = 0

By substituting the value of ( L+N+P ) IN ( 2)

=> Ꮮ+Ꮇ+N+Ꮲ+Ꭱ=0

=> Ꮇ + Ꭱ= 0

ᎻᎬNᏟᎬ ᏢᎡᎾᏙᎬᎠ !!

ᎻᎾᏢᎬ ᏆᎢ ᏔᏆᏞᏞ ᎻᎬᏞᏢFᏌᏞ ☺️

Answer 2

Hey Mate !

Here is your solution :

Given,

G( x ) = Lx^4 + Mx^3 + Nx^2 + Rx + P

It is given that ( 1 ) and ( -1 ) are its zeroes.

So,

=> x = 1 , ( -1 )

When,

=> x = 1

By Remainder Theorem ,

G( x ) = Lx^4 + Mx^3 + Nx^2 + Rx + P = 0

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

=> L × 1 + M × 1 + N × 1 + R + P = 0

=> L + M + N + R + P = 0 ----------- ( 1 )

When,

=> x = -1

By Remainder Theorem ,

=> G( x ) = Lx^4 + Mx^3 + Nx^2 + Rx + P = 0

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

=> L× 1 + M( -1 ) + N ( 1 ) - R + P = 0

=> L - M + N - R + P = 0 ------- ( 2 )

Adding ( 1 ) and ( 2 ),

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

=> L - M + N - R + P = 0

-------------------------------------

=> 2L + 2N + 2P = 0

=> 2 ( L + N + P ) = 0

=> ( L + N + P ) = 0 ÷ 2

=> L + N + P = 0 -------- ( 3 )

By substituting the value of ( 3 ) in ( 1 ),

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

=> ( L + N + P ) + M + R = 0

=> 0 + ( M + R ) = 0

=> ( M + R ) = 0 - 0

=> ( M + R ) = 0 --------- ( 4 )

From ( 3 ) and ( 4 ),

=> ( L + N + P ) = ( M + R ) = 0

★ Proved ★

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Hope it helps !! ^_^