If the expression x² – 2ax + a² leaves a remainder 4 when it is divided by x +…

Remainder Theorem: dividing a polynomial p(x) by a linear factor (x − c) leaves a remainder equal to p(c) — the polynomial evaluated at x = c.

Applying this to the given polynomial:

1. The divisor x + a can be written as x − (−a), so c = −a.

2. Substitute x = −a into x2 − 2ax + a2: p(−a) = (−a)2 − 2a(−a) + a2 = a2 + 2a2 + a2 = 4a2.

3. The remainder is given as 4, so 4a2 = 4, giving a2 = 1.

4. Taking the square root of both sides gives a = ±1.

Cross-check: x2 − 2ax + a2 is the perfect square (x − a)2. Since 4a2 depends only on a2, substituting a = 1 or a = −1 gives the identical result — dividing x2 − 2x + 1 by x + 1 leaves remainder (−1)2 − 2(−1) + 1 = 1 + 2 + 1 = 4, confirming the remainder condition holds for both a = 1 and a = −1, i.e. a = ±1.