LimpideTOOLS · VERIFIED EXERCISES

Special products.

Three formulas usually learnt by heart. With a figure for each, there is nothing left to learn by heart.

PRACTISEAn exercise on the identities

A perfect square; the tool chooses the seed.

IDENTITY 1 · SQUARE OF A SUMTRAP · (a + b)² ≠ a² + b²
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The square of a sum:

(a+b)2=a2+2ab+b2{(a + b)}^{2} = a^{2} + 2ab + b^{2}
FIGURE · A SQUARE WITH SIDE a + bFour areas, four terms.
A square with side a + b, cut into a square a², a square b² and two rectangles ab.a²ababb²abab

WHY IT IS TRUE

  1. Read the figure: the large square is the sum of the four areas.

    (a+b)2=a2+ab+ab+b2{(a + b)}^{2} = a^{2} + ab + ab + b^{2}

  2. Or expand, multiplying each term of the first bracket by the second:

    (a+b)(a+b)=a2+ab+ba+b2{(a + b)}{(a + b)} = a^{2} + ab + ba + b^{2}

  3. The two cross products are equal: group them.

    (a+b)2=a2+2ab+b2{(a + b)}^{2} = a^{2} + 2ab + b^{2}

    The trap: (a + b)² ≠ a² + b². The two ab rectangles of the figure are exactly what the error leaves out.

  4. Read it backwards, to factor: two squares, a² and b², and in the middle twice the product of a and b, 2ab, after a plus or a minus; that sign is the one in the bracket. In x² + 6x + 9, a = x and b = 3, and 6x = 2 · x · 3: it is (x + 3)².

    a2+2ab+b2=(a+b)2a^{2} + 2ab + b^{2} = {(a + b)}^{2}

WITH NUMBERS

(x+3)2=x2+6x+9{(x + 3)}^{2} = x^{2} + 6x + 9

a = x and b = 3: twice the product is 2 · x · 3 = 6x.

Practise

The tool opens a factoring exercise on this identity.

IDENTITY 2 · SQUARE OF A DIFFERENCETRAP · (a − b)² ≠ a² − b²
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The square of a difference:

(a−b)2=a2−2ab+b2{(a - b)}^{2} = a^{2} - 2ab + b^{2}
FIGURE · THE SAME SQUARE, CUT ANOTHER WAYTwo strips taken away, one corner given back.
A square with side a, from which two strips ab of width b are taken away; they overlap on the corner b². What is left is the square (a − b)².(a − b)²ababb²a − bba − bba

WHY IT IS TRUE

  1. From the square with side a, take away the two strips of width b:

    a2−ab−ab=a2−2aba^{2} - ab - ab = a^{2} - 2ab

  2. The two strips overlap on the corner b²: taken away twice, it must be given back once.

    (a−b)2=a2−2ab+b2{(a - b)}^{2} = a^{2} - 2ab + b^{2}

    The trap: (a − b)² ≠ a² − b². The b² comes back with a plus, because it was taken away twice.

  3. By calculation, it is the first identity with −b in place of b:

    (a+(−b))2=a2+2a(−b)+(−b)2{(a + {(-b)})}^{2} = a^{2} + 2a{(-b)} + {(-b)}^{2}

  4. Read it backwards, to factor: two squares, a² and b², and in the middle twice the product of a and b, 2ab, after a plus or a minus; that sign is the one in the bracket. In 4x² − 20x + 25, a = 2x and b = 5, and 20x = 2 · 2x · 5: it is (2x − 5)².

    a2−2ab+b2=(a−b)2a^{2} - 2ab + b^{2} = {(a - b)}^{2}

WITH NUMBERS

(2x−5)2=4x2−20x+25{(2x - 5)}^{2} = 4x^{2} - 20x + 25

a = 2x and b = 5: twice the product is 2 · 2x · 5 = 20x. The square of 2x is 4x², not 2x².

Practise

The tool opens a factoring exercise on this identity.

IDENTITY 3 · DIFFERENCE OF SQUARESTRAP · a² − b² ≠ (a − b)²
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A difference of squares:

(a+b)(a−b)=a2−b2{(a + b)}{(a - b)} = a^{2} - b^{2}
FIGURE · THE L SHAPE BECOMES A RECTANGLESame area, two shapes.
On the left, the L shape left when the square b² is taken from the corner of the square a², cut into two rectangles a(a − b) and b(a − b); on the right, the two rectangles side by side, with sides a + b and a − b.b(a − b)a(a − b)b²a − bba=a(a − b)b(a − b)aba − b

WHY IT IS TRUE

  1. Take the square b² from the corner of the square a²: an L shape is left.

    a2−b2a^{2} - b^{2}

  2. Cut the L shape into two rectangles and lay them side by side:

    a(a−b)+b(a−b)a{(a - b)} + b{(a - b)}

  3. The rectangle has sides a + b and a − b: factor out (a − b).

    a2−b2=(a+b)(a−b)a^{2} - b^{2} = {(a + b)}{(a - b)}

    The trap: a² − b² ≠ (a − b)². A difference of squares is not the square of a difference.

  4. By calculation, expanding: the cross terms −ab and ab cancel out.

    (a+b)(a−b)=a2−ab+ab−b2{(a + b)}{(a - b)} = a^{2} - ab + ab - b^{2}

  5. To factor, recognise this form: two squares separated by a minus. In x² − 16, a = x and b = 4:

    x2−16=(x+4)(x−4)x^{2} - 16 = {(x + 4)}{(x - 4)}

WITH NUMBERS

(x+4)(x−4)=x2−16{(x + 4)}{(x - 4)} = x^{2} - 16

a = x and b = 4: the cross terms 4x and −4x cancel out.

Practise

The tool opens a factoring exercise on this identity.