Q. \( f(x) = s(x) (x^2 – 4) + t(x). \)

Solution

We are given the identity

\(f(x)=s(x)\bigl(x^{2}-4\bigr)+t(x)\)

and asked to explain this step by step. I will show why such polynomials \(s(x)\) and \(t(x)\) always exist and how to determine the remainder polynomial \(t(x)\) explicitly. I will give the detailed algebraic steps and the method to obtain \(s(x)\) if required.

Step 1 — Division algorithm for polynomials (existence and form of the remainder)

By the polynomial division algorithm, for any polynomial \(f(x)\) and any nonzero polynomial \(g(x)\) there exist unique polynomials \(q(x)\) and \(r(x)\) such that

\(f(x)=q(x)g(x)+r(x)\)

with either \(r(x)=0\) or \(\deg r(x)<\deg g(x)\). Here take \(g(x)=x^{2}-4\), which has degree 2. Therefore there exist unique polynomials \(s(x)\) and \(t(x)\) such that

\(f(x)=s(x)\bigl(x^{2}-4\bigr)+t(x)\)

with \(\deg t(x)<2\). Thus \(t(x)\) must be at most linear, so we may write

\(t(x)=ax+b\)

for some constants (or coefficients) \(a\) and \(b\) to be determined from \(f\).

Step 2 — Evaluate at the roots of \(x^{2}-4\)

The polynomial \(x^{2}-4\) factors as \((x-2)(x+2)\). Its roots are \(x=2\) and \(x=-2\). At these values the product \(s(x)\bigl(x^{2}-4\bigr)\) vanishes, so the identity simplifies and gives direct equations for \(t\).

Evaluate the identity at \(x=2\):

\(f(2)=s(2)\bigl(2^{2}-4\bigr)+t(2)=0+t(2)=t(2)\).

Since \(t(2)=a\cdot 2+b\), we obtain

\(f(2)=2a+b.\)

Evaluate the identity at \(x=-2\):

\(f(-2)=s(-2)\bigl((-2)^{2}-4\bigr)+t(-2)=0+t(-2)=t(-2)\).

Since \(t(-2)=a\cdot(-2)+b\), we obtain

\(f(-2)=-2a+b.\)

Step 3 — Solve the linear system for a and b

We now have the linear system

\(2a+b=f(2)\)

\(-2a+b=f(-2)\)

Subtract the second equation from the first to eliminate \(b\):

\((2a+b)-(-2a+b)=f(2)-f(-2)\)

\(4a=f(2)-f(-2)\)

Therefore

\(a=\dfrac{f(2)-f(-2)}{4}.\)

Add the two equations to eliminate \(a\):

\((2a+b)+(-2a+b)=f(2)+f(-2)\)

\(2b=f(2)+f(-2)\)

Therefore

\(b=\dfrac{f(2)+f(-2)}{2}.\)

Step 4 — Write the explicit remainder \(t(x)\)

Substitute the values of \(a\) and \(b\) into \(t(x)=ax+b\). The remainder polynomial is

\(t(x)=\dfrac{f(2)-f(-2)}{4}\,x+\dfrac{f(2)+f(-2)}{2}.\)

This expression gives the unique polynomial of degree less than 2 that satisfies the original identity. It depends only on the values of \(f\) at \(x=2\) and \(x=-2\).

Step 5 — How to get \(s(x)\) if you need it

Once \(t(x)\) is known, the quotient \(s(x)\) is obtained by polynomial division of \(f(x)-t(x)\) by \(x^{2}-4\). Concretely, compute

\(s(x)=\dfrac{f(x)-t(x)}{x^{2}-4}\)

and perform polynomial long division (or synthetic division twice using the linear factors \(x-2\) and \(x+2\)) to obtain the polynomial \(s(x)\). The division will produce a polynomial because \(f(x)-t(x)\) is divisible by \(x^{2}-4\) by construction.

Conclusion

For any polynomial \(f(x)\) there exist unique polynomials \(s(x)\) and \(t(x)\) with \(\deg t<2\) such that

\(f(x)=s(x)\bigl(x^{2}-4\bigr)+t(x)\).

The remainder \(t(x)\) is explicitly

\(t(x)=\dfrac{f(2)-f(-2)}{4}\,x+\dfrac{f(2)+f(-2)}{2}.\)

If required, compute \(s(x)\) by dividing \(f(x)-t(x)\) by \(x^{2}-4\) using polynomial long division or by successive synthetic division by \(x-2\) and \(x+2\).