Factorise \(100a^2-b^2\).

Correct answer: \((10a-b)(10a+b)\)

The direct answer is A: \((10a-b)(10a+b)\). Recognize a difference of squares: \(100a^2=(10a)^2\) and \(b^2=b^2\). Apply \(P^2-Q^2=(P-Q)(P+Q)\) with \(P=10a\) and \(Q=b\). This gives \(100a^2-b^2=(10a-b)(10a+b)\). Expanding confirms it: the cross terms \(+10ab\) and \(-10ab\) cancel, leaving \(100a^2-b^2\). B is not the correct conjugate pair and expands to \(100a^2+99ab-b^2\). C is a square of one binomial and expands to \(100a^2-20ab+b^2\), with extra terms and the wrong sign on \(b^2\). D treats \(100a\), rather than its square root \(10a\), as the factor base; it gives \(10000a^2-b^2\). Memory cue: take square roots first, then use one minus and one plus factor.