Visual models of identities

These sides give parts (3x^2), (9x), (x) and (3). In exams, check by adding all four parts.

The difference is ((x+6+x+2)(x+6-x-2)= (2x+8)\cdot4=8x+32). In exams, use difference of squares.

The area difference is ((x+5)^2-(x-1)^2=12x+24). In exams, subtract inner square from outer square.

Area is unchanged and one side is unchanged, so the other side is also (a+b). In exams, infer using equal area and equal side.

The uncolored area is two (3x) strips, so (6x=30) and (x=5). In exams, connect uncolored strips to the middle term.

From \((x-4)^2=81\), \(x-4=9\) and \(x=13\). In exams, treat the side as a positive length.

The first area is (x^2+7x+10) and the second is (x^2+7x+12). In exams, when linear terms match, compare constants.

One side of the large rectangle is \(a+b\) and the other is \(c+d\). The four smaller rectangles therefore have areas \(ac\), \(ad\), \(bc\), and \(bd\). Hence, \((a+b)(c+d)=ac+ad+bc+bd\). Option A incorrectly includes terms such as \(ab\) and \(cd\), which do not come from multiplying one part of each different side. Exam tip: in an area model, form every term by multiplying one segment from the first side by one segment from the second side.

( (x+5)^2=x^2+10x+25 ) and ( (x+2)(x+8)=x^2+10x+16 ). In exams, cancel equal (x) terms and compare constants.

It directly shows the four parts of ( (x+k)^2 ). In exams, view (2kx) as two (kx) strips.

Two (kx) strips are subtracted and the (k^2) corner is added back. In exams, identify (x-k) from (-2kx).

Combining two equal rectangles gives (2xy). In exams, rearrangement does not change area.

Among the three mixed pairs, the (ca) pair also appears twice. In exams, complete all pair combinations.

( (x+4)^2=x^2+8x+16 ), so removing the corner leaves (x^2+8x). In exams, subtract the removed part.

The total of two mixed rectangles is (2\cdot2x\cdot7=28x). In exams, find middle area with coefficients.

The linear parts are (-5x) and (-7x), whose sum is (-12x). In exams, add both negative strips.

(-5+(-7)=-12) and ((-5)(-7)=35). In exams, product of two negative constants is positive.

The corner (b^2) is removed twice, so it must be added back. In exams, correct over-subtraction.

The two square areas are \\((x+k)^2\\) and \\((x-k)^2\\). Expanding them gives \\(x^2+2kx+k^2\\) and \\(x^2-2kx+k^2\\). When the second area is subtracted from the first, the equal terms \\(x^2\\) and \\(k^2\\) cancel. The remaining terms are \\(2kx-(-2kx)=4kx\\), so the difference is \\(4kx\\).

Option A is correct. In a visual area model, the two squares share the same main square contribution and the same small square contribution; only the cross-rectangles create the net difference. A common error is choosing \\(2kx\\) by noticing only one cross-term, but each expansion contains a cross-term of opposite sign, so together they contribute \\(4kx\\). Thus the identity \\((x+k)^2-(x-k)^2=4kx\\) confirms A.

Both have (x^2+(a+b)x) common, and the constant difference is ( \frac{(a+b)^2}{4}-ab=\frac{(a-b)^2}{4} ). In exams, remove common parts and compare remaining area.