Question

the remainder of the polynomial 5+bx-2x2+ax3,when divided by x-2 is twice the remainder when it is divided by (x+1).show that 10a+4b=9.​

Answer 1

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Answer 2

Below it is proved that 10a+4b=9.

Step-by-step explanation:

The given polynomial is

$P(x)=5+bx-2x^2+ax^3$

According to remainder theorem, if P(x) is divided by (x-c), then remainder is P(c).

Substitute x=2 in the given polynomial.

$P(2)=5+b(2)-2(2)^2+a(2)^3=8a+2b-3$

Substitute x=-1 in the given polynomial.

$P(-1)=5+b(-1)-2(-1)^2+a(-1)^3=-a-b+3$

It is given that the remainder of the polynomial P(x), when divided by x-2 is twice the remainder when it is divided by (x+1).

$P(2)=2\times P(-1)$

$8a+2b-3=2(-a-b+3)$

$8a+2b-3=-2a-2b+6$

$8a+2b+2a+2b=3+6$

$10a+4b=9$

Hence proved.

#Learn more

A polynomial ax3 - 3x2 + 3x + 1 when divided by X-2 leaves the remainder 7. Find the vale of a.

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