Can't use $a^2 - b^2$

Algebra Level 4

Let $a$ be a non-zero real number and $p(x)$ be a polynomial of degree greater than 2. If $p(x)$ leaves remainder $a$ and $-a$ respectively by $x+a$ and $x-a$ , the remainder when $p(x)$ is divided by $x^2 - a^2$ is

-2x

x

2x

-x

1 solution

Department 8
Oct 1, 2015

I am taking $p(x)=f(x)$ .

We have (can be proved using remainder theorem)

$f\left( a \right) =-a \\ f\left( -a \right) =a$

When $f\left( x \right)$ is divided by $(x+a)(x-a)$ , let the quotient be $g(x)$ and remainder be $bx+c$ .

$f\left( x \right) =\left( x-a \right) \left( x+a \right) g\left( x \right) +bx+c$

Keeping $x=a,-a$ . We will have

$-a=ab+c\\ a=-ab+c\\ \therefore c=0,b=-1$

The remainder was $bx+c$ . Keeping the values of $(b,c)$ , then we have remainder as $-x$

I think you should also have explained why you have taken the remainder as a linear polynomial.

Manish Mayank - 5 years, 8 months ago

Can you give me something to copy because I am not able to think what to right.

Department 8 - 5 years, 8 months ago

What....!!??

Manish Mayank - 5 years, 8 months ago

@Manish Mayank – Yes, for the suggestion you gave on the comment why the remainder is a linear polynomial.

Department 8 - 5 years, 8 months ago

@Department 8 – Actually I am not quite sure of that.. :(

Manish Mayank - 5 years, 8 months ago

Can't use a 2 − b 2 a^2 - b^2 a 2 − b 2

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Can't use $a^2 - b^2$