Visual models of identities

Since (5+6=11) and (5\cdot6=30), the strips are (5x) and (6x). In exams check both conditions together.

Difference of squares is shown by subtracting the smaller square from the larger square. In exams think of (a^2-b^2) as a cut-out area.

Since (81=9^2) and (2\cdot x\cdot9=18x), the side is (x+9). In exams take the square root of the constant square.

The total area is \(a^2+ab+ab+b^2=a^2+2ab+b^2\), equal to the area of a square with side \(a+b\). Option B has a negative middle term. Exam tip: two \(ab\) rectangles indicate \(+2ab\).

((a+3)(a-3)=a^2-9). In exams when the same number appears with (+) and (-), write difference of squares.

(25x^2=(5x)^2) and (36=6^2), so it is a difference of squares. In exams first identify the sides of both squares.

The areas of the two squares are \((x+1)^2\) and \((x-1)^2\), respectively. Their difference is \((x+1)^2-(x-1)^2=(x^2+2x+1)-(x^2-2x+1)=4x\). \(x^2-1\) equals \((x+1)(x-1)\), not the difference of the areas. Exam tip: use \(a^2-b^2=(a-b)(a+b)\) to find such differences quickly.

The square has two (10x) strips and a (100) corner. In exams looking at four regions separately reduces mistakes.

The total height is \(x+8\) and the total width is \(x+2\). Hence, the area is \((x+8)(x+2)=x^2+2x+8x+16=x^2+10x+16\). In option B, the coefficient of the \(x\)-term and the constant term are incorrectly interchanged. Exam tip: add the areas of the four smaller rectangles: \(x^2, 2x, 8x\), and \(16\).

Since (4\cdot9=36) and (4+9=13), these are the correct sides. In exams both product and sum must match.

(x^2-25=(x+5)(x-5)). In exams think of rearranging the remaining area after removing the small square.

(9a^2=(3a)^2), (16=4^2), and (2\cdot3a\cdot4=24a). In exams the side is made from the roots of the two squares.

Dividing a square of side \((4x-1)\) into parts \(4x\) and \(-1\) gives two rectangles. Each rectangle contributes \(-4x\), so their combined area, the middle term, is \(-4x-4x=-8x\). Hence, \((4x-1)^2=16x^2-8x+1\). \(8x\) has the wrong sign because the side contains a negative term. Exam tip: in \((a-b)^2\), the middle term is always \(-2ab\).

((x-3)(x-7)=x^2-10x+21), and the parts show this. In exams two negative strips create a positive corner.

The removed strips are (4x) and (9x), and the corner is (4\cdot9=36). In exams keep unequal removed strips separate.

Since (3+(-5)=-2) and (3\cdot(-5)=-15), the sides are (x+3) and (x-5). In exams check sum and product with signs.

The difference in areas is \((x+6)^2-(x-6)^2\). Using \(a^2-b^2=(a-b)(a+b)\), we get \(((x+6)-(x-6))((x+6)+(x-6))=12\times 2x=24x\). In the visual model, this can be divided into four rectangular strips, each of area \(6x\), giving a total of \(4\times 6x=24x\). The \(36\)-area corner parts cancel or adjust in the arrangement, so four \(36\) squares alone are not the result. Exam tip: when subtracting areas of two squares, first look for the identity \(a^2-b^2\).

In ((a+b)^2), the two middle strips are (ab) and (ab). In exams look for equal strips in a perfect square.

Two (5x) strips were missed, so total (10x) was missed. In exams do not treat ((a+b)^2) as (a^2+b^2).

When two strips are removed, the corner is removed twice, so (64) is added back. In exams remember the added last term in a subtraction square.