When a certain polynomial is divided by x - 2, the remainder is 2. When the polynomial is divided by x + 2, the remainder is -2. What is the remainder when the polynomial is divided by x^2 - 4?

Guest Dec 19, 2019

#1**+1 **

Let's say the certain polynomial is \(f(x)\). Then

\(f(x)=g_1(x)\cdot(x-2)+2\), and

\(f(x)=g_2(x)\cdot(x+2)-2\).

so

\(f(2)=g_1(2)\cdot(2-2)+2=2\), and

\(f(-2)=g_2(-2)\cdot(-2+2)-2=-2\)

If we divide \(f(x)\) by \(x^2-4\), we end up with a quotient polynomial \(g(x)\) and remainder \(r(x)\),

where \(r(x)\) is a linear function of x, since it's degree is1 less that the degree of the divisor polynomial. So

we can write

\(f(x)=g(x)\cdot(x^2-4)+r(x)\), where \(r(x)=mx+b\).

Now evaluate the function \(f\) at 2 and -2:

\(f(2)=g(2)\cdot(2^2-4)+r(2)=0+r(2)=2\), which implies

\(r(2)=2m+b=2\), and

\(f(-2)=g(-2)\cdot((-2)^2-4)+r(-2)=r(-2) \), which means

\(r(-2)=-2m+b=-2\).

By eliminating b from the above two equations we get \(m=1 \) and \(b=0\); that gives us the remainder

\(r(x)=x\)

.Gadfly Dec 19, 2019