ICSE Class 7 Maths Chapter 14 Products

Chapter 14: Products

Some Special Products As Identity

\((x + a)(x + b) = x(x + b) + a(x + b)\)

\(= x^2 + bx + ax + ab\)

\(= x^2 + ax + bx + ab = x^2 + (a + b)x + ab\)

Consider, \((x + 5)(x + 2) = x^2 + (5 + 2)x + (5 \times 2)\)

\(= x^2 + 7x + 10\)

If \(x = 3\), it becomes:

\((3 + 5)(3 + 2) = 3^2 + 7 \times 3 + 10\)

\(8 \times 5 = 9 + 21 + 10\)

\(40 = 40\); which is true.

For \(x = -8\), we get:

\((-8 + 5)(-8 + 2) = (-8)^2 + 7 \times (-8) + 10\)

\(-3 \times (-6) = 64 - 56 + 10\)

\(18 = 18\); which is true.

Similarly, for each and every value of \(x\), we shall find \((x + 5)(x + 3) = x^2 + 7x + 10\) is true.

As such a statement, which is true for each and every value of its variable, is called an Identity.

In the same way, each of the following is also an identity.

\((x + a)(x - b) = x(x - b) + a(x - b)\)

\(= x^2 - bx + ax - ab\)

\(= x^2 + ax - bx - ab = x^2 + (a - b)x - ab\)

\((x - a)(x + b) = x(x + b) - a(x + b)\)

\(= x^2 + bx - ax - ab\)

\(= x^2 - ax + bx - ab = x^2 - (a - b)x - ab\)

\((x - a)(x - b) = x(x - b) - a(x - b)\)

\(= x^2 - bx - ax + ab\)

\(= x^2 - ax - bx + ab = x^2 - (a + b)x + ab\)

Hence,

\((x + a)(x + b) = x^2 + (a + b)x + ab\)

\((x + a)(x - b) = x^2 + (a - b)x - ab\)

\((x - a)(x + b) = x^2 - (a - b)x - ab\)

\((x - a)(x - b) = x^2 - (a + b)x + ab\)

It can be combined as:

\((x + a)(x + b) = x^2 + \text{(sum of second terms with their given signs)}x + \text{(product of second terms)}\).

Examples

\((x + 10)(x + 7) = x^2 + (10 + 7)x + (10 \times 7)\)

\(= x^2 + 17x + 70\)

\((x + 10)(x - 7) = x^2 + (10 - 7)x + (10 \times (-7))\)

\(= x^2 + 3x - 70\)

\((x - 10)(x + 7) = x^2 + (-10 + 7)x + (-10 \times 7)\)

\(= x^2 - 3x - 70\)

\((x - 10)(x - 7) = x^2 + (-10 - 7)x + (-10 \times (-7))\)

\(= x^2 - 17x + 70\)

In the same way:

\((y + 15)(y - 3) = y^2 + (15 - 3)y + (15 \times (-3))\)

\(= y^2 + 12y - 45\)

\((m - 9)(m + 4) = m^2 + (-9 + 4)m + (-9 \times 4)\)

\(= m^2 - 5m - 36\)

\((p - \frac{1}{2})(p + \frac{3}{4}) = p^2 + (-\frac{1}{2} + \frac{3}{4})p + (-\frac{1}{2} \times \frac{3}{4})\)

\(= p^2 + \frac{1}{4}p - \frac{3}{8}\)

\((n + 0.6)(n + 0.8) = n^2 + (0.6 + 0.8)n + (0.6 \times 0.8)\)

\(= n^2 + 1.4n + 0.48\)

Teacher's Note

When you buy items at a store with a discount on top of a promotion, you are using algebraic products to calculate the final price - these formulas help us save time with repeated calculations in real life.

Exercise 14(A)

1. Find:

(i) \((x + 5)(x + 12)\) (ii) \((x + 8)(x + 2)\) (iii) \((y + 3)(y + 15)\)

(iv) \((n + 12)(n + 15)\) (v) \((x + 0.4)(x + 0.5)\) (vi) \((y + 0.7)(y + 0.1)\)

(vii) \((x + \frac{1}{3})(x + \frac{2}{3})\) (viii) \((y + \frac{4}{5})(y + \frac{2}{5})\)

2. Find:

(i) \((x + 4)(x - 1)\) (ii) \((x + 15)(x - 8)\) (iii) \((m + 8)(m - 8)\)

(iv) \((n + 15)(n - 12)\) (v) \((x + 0.8)(x - 0.5)\) (vi) \((y + 0.7)(y - 0.1)\)

(vii) \((x + \frac{3}{4})(x - \frac{1}{4})\) (viii) \((y + \frac{4}{5})(y - \frac{2}{5})\)

3. Find:

(i) \((x - 8)(x + 5)\) (ii) \((x - 12)(x + 7)\) (iii) \((m - 5)(m + 5)\)

(iv) \((x - 18)(x + 3)\) (v) \((x - 1.6)(x + 0.9)\) (vi) \((y - 0.8)(y + 0.1)\)

(vii) \((x - \frac{7}{15})(x + \frac{2}{15})\) (viii) \((y - \frac{4}{5})(y + \frac{3}{5})\)

4. Find:

(i) \((x - 7)(x - 4)\) (ii) \((x - 5)(x - 20)\) (iii) \((m - 9)(m - 9)\)

(iv) \((n - 18)(n - 8)\) (v) \((x - 0.7)(x - 1.4)\) (vi) \((y - 2.4)(y - 0.6)\)

(vii) \((x - \frac{4}{9})(x - \frac{5}{9})\) (viii) \((y - \frac{7}{13})(y - \frac{3}{13})\)

5. Find:

(i) \((x + 15)(x + 9)\) (ii) \((x + 15)(x - 9)\) (iii) \((x - 15)(x + 9)\)

(iv) \((x - 15)(x - 9)\) (v) \((y + 0.8)(y + 0.4)\) (vi) \((y + 0.8)(y - 0.4)\)

(vii) \((y - 0.8)(y + 0.4)\) (viii) \((y - 0.8)(y - 0.4)\) (ix) \((z + \frac{2}{5})(z - \frac{1}{5})\)

(x) \((z - \frac{2}{5})(z - \frac{1}{5})\) (xi) \((z - \frac{2}{5})(z + \frac{1}{5})\)

Square Of The Sum Of Two Terms

\((a + b)^2 = (a + b)(a + b)\)

\(= a(a + b) + b(a + b)\)

\(= a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\)

Thus: \((a + b)^2 = a^2 + 2ab + b^2\)

(Sum of two terms)^2 = (1st term)^2 + 2 \times 1st term \times 2nd term + (2nd term)^2

Using the same property (formula), we get:

\((2x + y)^2 = (1st \text{ term})^2 + 2 \times 1st \text{ term} \times 2nd \text{ term} + (2nd \text{ term})^2\)

\(= (2x)^2 + 2 \times 2x \times y + (y)^2\)

\(= 4x^2 + 4xy + y^2\)

\((3m + 5n)^2 = (3m)^2 + 2 \times 3m \times 5n + (5n)^2\)

\(= 9m^2 + 30mn + 25n^2\)

Square of \(5a + \frac{2}{3}b = \left(5a + \frac{2}{3}b\right)^2\)

\(= (5a)^2 + 2 \times 5a \times \frac{2}{3}b + \left(\frac{2}{3}b\right)^2\)

\(= 25a^2 + \frac{20}{3}ab + \frac{4}{9}b^2\)

Square Of The Difference Of Two Terms

\((a - b)^2 = (a - b)(a - b)\)

\(= a(a - b) - b(a - b)\)

\(= a^2 - ab - ba + b^2 = a^2 - 2ab + b^2\)

Thus: \((a - b)^2 = a^2 - 2ab + b^2\)

(Difference of two terms)^2 = (1st term)^2 - 2 \times 1st term \times 2nd term + (2nd term)^2.

Using the same property (formula), we get:

\((3x - 2y)^2 = (1st \text{ term})^2 - 2 \times 1st \text{ term} \times 2nd \text{ term} + (2nd \text{ term})^2\)

\(= (3x)^2 - 2 \times 3x \times 2y + (2y)^2\)

\(= 9x^2 - 12xy + 4y^2\)

\((5m - 4n)^2 = (5m)^2 - 2 \times 5m \times 4n + (4n)^2\)

\(= 25m^2 - 40mn + 16n^2\)

Square of \(3x - \frac{5}{6}y = \left(3x - \frac{5}{6}y\right)^2\)

\(= (3x)^2 - 2 \times 3x \times \frac{5}{6}y + \left(\frac{5}{6}y\right)^2 = 9x^2 - 5xy + \frac{25}{36}y^2\)

Teacher's Note

Perfect squares appear in construction when you need to find the area of a square-shaped room - knowing the formula helps you calculate quickly without measuring.

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