Question

if the polynomial 2x cube+ax square+3x-5and x cube +2x square -5x -a leaves the same remainder when divided by x-1 then find the value of a
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Answer 1

Answer:

a= -1

Step-by-step explanation:

p(x)=2x^3+ax^2+3x-5

p(x) is divided by x-1

Remainder = 2(1)^3+ a(1)^2+3×1-5

=2+a+3-5

= a

Again, q(x) = x^3 + 2x^2 - 5x - a

Since, q(x) is divided by x-1 then

Remainder = (1)^3 + 2(1)^2 - 5× 1- a

= 1+2-5-a

= -2-a

A/Q

Since, Both remainder are equal

Therefore, a = -2-a

2a= -2

a = -1.

Answer 2

Answer:

a=-1

P(x)-2x3tax^2+3x-5

p(x) is divided by x-1

Remainder 2(13+ a(1)^2+3x1-5

=2ta+3-5

a

Again, q(x) = x^3 + 2x^2 - 5x - a

Since, qx) is divided by x-1 then

Remainder = (13+ 2(1^2 5x 1- a

= 1+2-5-a

-2-a

A/Q

Since, Both remainder are equal

Therefore, a = -2-a

2a-2

a = -1.

Step-by-step explanation:

Hope it helps you

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