Visual models of identities

The area of the square is \((a+b)^2\) and its parts give \(a^2+2ab+b^2\). Exam tip: split the big square into smaller parts.

The square with side \(a\) has area \(a^2\), and the square with side \(b\) has area \(b^2\). Each rectangle has sides \(a\) and \(b\), so its area is \(ab\). The two rectangles together have area \(2ab\). Hence, the total area is \(a^2+2ab+b^2\). Option A leaves out the areas of the rectangles. Exam tip: Add the areas of all the separate parts in a visual model.

The area is \((x+3)^2\) and \((a+b)^2\) gives \(x^2+6x+9\). Exam tip: do not miss the middle term \(2ab\).

The smaller square is subtracted from the larger square so we get \(a^2-b^2\). Exam tip: remember to subtract the removed part.

The difference of squares model shows \(a^2-b^2=(a-b)(a+b)\). Exam tip: look for both difference and sum factors.

The area of a square is (side)². Therefore, for side \(a-b\), its area is \((a-b)^2\). Using the identity \((a-b)^2=a^2-2ab+b^2\), the correct answer is \(a^2-2ab+b^2\). The expression \(a^2-b^2\) equals \((a-b)(a+b)\), so it is not the area of this square. Exam tip: in \((a-b)^2\), the middle term is always \(-2ab\).

Two \(ab\) areas are subtracted and \(b^2\) is added back once. So \((a-b)^2=a^2-2ab+b^2\).

Dividing the rectangle into four smaller parts gives areas \(x\cdot x=x^2\), \(x\cdot 5=5x\), \(2\cdot x=2x\), and \(2\cdot 5=10\). Therefore, the total area is \(x^2+5x+2x+10=x^2+7x+10\). Option A has 7 instead of the required constant term 10, so it is incorrect. Exam tip: when multiplying two binomials, remember to add both middle terms.

The tiles represent \(x^2+8x+16\). Since \(x^2+8x+16=x^2+2\cdot4\cdot x+4^2\), it is \((x+4)^2\). In contrast, \((x+8)^2\) would contain the middle term \(16x\), not \(8x\). Exam tip: for \(x^2+bx+c\), if \(c=a^2\) and \(b=2a\), write it as \((x+a)^2\).

The small square has area \(25\), so its side is \(5\). In the square model, the two middle rectangles have total area \(2 \times x \times 5=10x\), which matches the given information. Therefore, the side of the larger square is \(x+5\). If the side were \(x+10\), the middle rectangles would have total area \(20x\). Exam tip: find the side of the small square by taking the square root of its area.

In \((p+q)^2\), there are two equal rectangles \(pq\) and \(pq\). Exam tip: remember the middle term \(2pq\).

The four parts are \(m^2\), \(mn\), \(mn\), and \(n^2\). \(m^2-n^2\) belongs to another identity.

Area of a rectangle = length × breadth. Therefore, the area is \((a+b)(a-b)=a^2-b^2\), using the identity \((x+y)(x-y)=x^2-y^2\). Option C is the expansion of \((a-b)^2\), so it does not apply here. Exam tip: in the product of a sum and a difference, the middle terms cancel.

Here, \(7=8-1\) and \(9=8+1\). Apply the identity \((a-b)(a+b)=a^2-b^2\): \(8^2-1^2=64-1=63\). Therefore, the correct value is \(63\). \(64\) is only \(8^2\); subtracting \(1^2\) is necessary. Exam tip: the product of the numbers one less and one more than \(n\) is \(n^2-1\).

The side of the square is \(10+2=12\), so its area is \(12^2=144\). In the visual model, the square is split into a \(10\times10\) square, two \(10\times2\) rectangles, and a \(2\times2\) square: \(100+20+20+4=144\). In \(124\), one \(10\times2\) part is missing. Exam tip: always include the middle term \(2ab\) in \((a+b)^2=a^2+2ab+b^2\).

Add the areas of all four parts: \(6^2=36\), each \(6\times4\) rectangle has area \(24\), and \(4^2=16\). Thus, the total area is \(36+24+24+16=100\). This is the square model of \((6+4)^2=10^2\). Choosing \(96\) would miss the area of one rectangle. Exam tip: add every region in the model, or use the identity \((a+b)^2\).

Use the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=100\) and \(b=1\), so \((100-1)^2=10000-200+1=9801\). Getting \(9901\) indicates an incorrect calculation of the subtracted middle term, \(2ab=200\). Exam tip: while squaring a number close to 100, make sure to subtract the middle term \(2ab\).

Use the identity \((x+a)^2=x^2+2ax+a^2\). Here, \(a^2=36\) gives \(a=6\), and \(2ax=2\times6x=12x\). Therefore, \(x^2+12x+36=(x+6)^2\). In contrast, \((x-6)^2\) has the middle term \(-12x\), so it is not correct. Exam tip: find the square root of the constant term first, then check the sign of the middle term.

The total area of the four parts is \(y^2+4y+4y+16=y^2+8y+16\). This is \(y^2+2\cdot y\cdot4+4^2\), so it equals \((y+4)^2\). In contrast, \((y-4)^2\) has the middle term \(-8y\), whereas this expression has \(+8y\). Exam tip: in a square area model, add the areas of the two equal rectangles to check the middle term.

\(s^2\) is the area of the small square whose side is \(s\). Exam tip: identify each part from its two sides.