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Exploring Algebraic Identities — Class 9 Mathematics

Ganita Manjari · Chapter 4

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Chapter summary

This chapter develops algebraic identities — equations true for all values of the variables — and shows how to verify them geometrically with squares, rectangles and cubes, and using algebra tiles. The basic square identities are and , extended to three terms by . The difference of squares is , and the product pattern leads to factorising quadratics by splitting the middle term. Cubic identities include , , the sum and difference of cubes , and . These identities speed up numerical squaring and cubing, support factorisation, and let us simplify rational algebraic expressions by cancelling common factors.

Identities vs equationsSquare identitiesCubic identitiesFactorisationSimplifying rational expressions

Key ideas & terms

Identity: An equation that is true for every value of the variables involved.
Equation: A statement that holds only for particular values of the variable, unlike an identity.
Factorisation: Writing an expression as a product of its factors, e.g. .
Difference of squares: The identity .
Splitting the middle term: Factorising by finding with and .
Algebra tiles: Square and rectangular pieces used to visualise products and factors of expressions.

Extra questions & answers

What is the difference between an equation and an identity?

An identity holds for all values of the variables, while an equation is true only for certain values.

Expand .

Using : .

Factorise .

It matches with , so it equals .

Use an identity to compute .

Write : .

State the identity for .

It is , the difference of squares.

Factorise by splitting the middle term.

Find : , so it equals .

Expand .

It equals .

What is the factorisation of ?

It is .

Find if .

Using the identity and gives .

How is a rational algebraic expression simplified?

By factorising numerator and denominator and cancelling common non-zero factors.

Think and Reflect

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Take any four consecutive integers . Their squares are . One pattern is for every . For example and . So the sum of the outer two squares exceeds the sum of the inner two squares by always. Another pattern can also be checked. Any such constant result can be proved by expanding with .

Think and Reflect

1. What can you say about and if ?

Since , the inequality gives i.e. so . Hence and must have opposite signs, one positive and one negative.

2. What can you say about and if ?

implies so and . Hence and have the same sign, both positive or both negative, and neither is zero.

3. When will be equal to ?

Equality requires so i.e. . Hence at least one of is zero.

Did you observe that and are both positive? What term will decide which is larger? Use the expansion of to decide.

Yes, both are squares and hence . From the extra term decides. If is positive the first is larger, if negative the second is larger, if zero they are equal.

Exercise Set 4.1

1. Using the identity , expand the following: (i)

Put . Then .

1. Using the identity , expand the following: (ii)

With , .

1. Using the identity , expand the following: (iii)

1. Using the identity , expand the following: (iv)

1. Using the identity , expand the following: (v)

. If read as it expands to .

1. Using the identity , expand the following: (vi)

2. Using the same identity, find the values of the following: (i)

. So .

2. Using the same identity, find the values of the following: (ii)

. So .

2. Using the same identity, find the values of the following: (iii)

. So .

Think and Reflect

What if we replace by in ?

Replacing by gives . This is also an identity true for all and is used for subtraction, e.g. computing .

Exercise Set 4.2

1. Factor completely: (i)

. So it is with . Hence .

1. Factor completely: (ii)

. So .

1. Factor completely: (iii)

. Hence .

1. Factor completely: (iv)

. So .

1. Factor completely: (v)

Take common factor : .

1. Factor completely: (vi)

Factor : .

2. Find the values of the following using the identity . (i)

. So .

2. Find the values of the following using the identity . (ii)

. So .

2. Find the values of the following using the identity . (iii)

. So .

Think and Reflect

Label the squares and rectangles in Fig. 4.4 so that it represents the identity .

The big square of side is partitioned into 9 pieces: three squares along the diagonal with areas (top-left, middle, bottom-right) and six rectangles: two of area (top-middle and middle-left), two of area (middle-right and bottom-middle), two of area (top-right and bottom-left). Adding all gives .

Exercise Set 4.3

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i)

Use or . . So . Alternatively gives same result.

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (ii)

. Use : .

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (iii)

. So .

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (iv)

. So .

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (v)

. So .

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (vi)

. Using : . Also gives same.

2. Factor using suitable identities: (i)

. So using .

2. Factor using suitable identities: (ii)

. Hence .

2. Factor using suitable identities: (iii)

This is because and .

2. Factor using suitable identities: (iv)

. So expression using .

2. Factor using suitable identities: (v)

Compare with . If the given expression is it is not a perfect square, but the intended perfect square form is . Generally use with signs.

3. Expand the following using the identity : (i)

3. Expand the following using the identity : (ii)

Write as .

4. Is this an identity?

Expand each: , , . Sum , which is in general. So it is not an identity. The correct identity is .

Think and Reflect

1. Try to evaluate the following using a suitable identity: (i)

Use as in Sridhara. .

1. Try to evaluate the following using a suitable identity: (ii)

1. Try to evaluate the following using a suitable identity: (iii)

1. Try to evaluate the following using a suitable identity: (iv)

. Pattern: numbers ending with 5 squared end with 25 and the leading part is e.g. with 25.

2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it. 2b 2c, a+b+c, a+b-c, a-b+c, a-b-c

The figures show type expansion leading to vs . It visualises and also that . The two rows together illustrate difference of squares identities.

Think and Reflect

Suppose is split as ; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Splitting as gave a rectangle by of unit squares which fits units. Splitting as would require a by array of unit squares to keep area , but then the side lengths would be and giving area . So it does not form a perfect rectangle with units; only the factor pair of whose sum is works.

Think and Reflect

1. Figure out the product of and using algebra tiles.

Represent horizontally and vertically. The big rectangle consists of one tile, -tiles () and unit tiles ( array). Hence .

2. Lay out algebra tiles for in such a way that you will see its factors.

We need two numbers with sum and product : and . Arrange one tile, -tiles split as on one side and on the other, and units in a array. The outer dimensions become by , so .

Think and Reflect

We have seen that . Also . Generalise the pattern to get an expression for .

In general . The coefficient of is sum of and constant term is product .

Now consider the case where we have a rectangle of sidelengths and , as shown in Fig. 4.8. What can you say about its area ? Fill in the blanks with the appropriate expressions to make the equation true. . Also, verify your answer using the distributive property.

. For : . Verification: . So blanks are .

Exercise Set 4.4

1. Fill in the blanks to complete the following identities: (i)

We need . Numbers and satisfy. So .

1. Fill in the blanks to complete the following identities: (ii)

Factor . Need numbers with product and sum : and . So . Hence blank is .

1. Fill in the blanks to complete the following identities: (iii)

. Check: . So blanks are and .

1. Fill in the blanks to complete the following identities: (iv)

Need factors of summing to : . So or .

2. Select and use the identity that will help you to find the following products without multiplying directly: (i)

Use : , so .

2. Select and use the identity that will help you to find the following products without multiplying directly: (ii)

. So . Or .

2. Select and use the identity that will help you to find the following products without multiplying directly: (iii)

Use with : .

2. Select and use the identity that will help you to find the following products without multiplying directly: (iv)

or . .

2. Select and use the identity that will help you to find the following products without multiplying directly: (v)

. So .

2. Select and use the identity that will help you to find the following products without multiplying directly: (vi)

No simple identity with same middle; directly . Using distributivity: .

2. Select and use the identity that will help you to find the following products without multiplying directly: (vii)

2. Select and use the identity that will help you to find the following products without multiplying directly: (viii)

As before .

3. Factor the following: (i)

Observe . The expression matches . So factor is .

3. Factor the following: (ii)

. So .

3. Factor the following: (iii)

Need numbers with sum and product : and . So .

3. Factor the following: (iv)

. So .

3. Factor the following: (v)

Check . So factor is .

Think and Reflect

James: Reshma: I have a different idea. I will find this product to get the answer. According to you, who is correct and why? Try to combine more such identities and find new results.

Both are correct. Multiplication is associative, and both use true identities — and . Expanding either way gives the same result: , and — identical. So both methods are valid, and combining identities this way leads to further results such as .

Think and Reflect

We already know that . Further, we have verified that . Observe that is a common factor of and . Do you think is also a factor of ? Note that . Can you see how is a factor of ? How about ? Does this also have as a factor?

Since , is indeed a factor. In general is divisible by for every natural . For so again is a factor. This can be proved by factor theorem: putting makes .

Think and Reflect

Try to simplify the following rational expression: (Hint: Factor and simplify the rational expressions assuming that ).

Factor denominator: need numbers sum product : and . So . Numerator can be grouped and cancelled with a common factor to leave a simplified fraction after cancelling or provided denominator non-zero.

Exercise Set 4.5

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (i)

Numerator . Denominator . These share no common factor, so in fully factored form the expression is .

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (ii)

Numerator . Denominator . So .

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (iii)

Numerator . Denominator . Cancelling : .

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (iv)

Numerator . Denominator . So for denominator .

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (v)

Factor each: , , , . Numerator and denominator are identical, so the expression simplifies to .

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (vi)

. Denominator . Cancelling one gives for .

End-of-Chapter Exercises

1. Use suitable identities to find the following products: (i)

using with and sign handling, or .

1. Use suitable identities to find the following products: (ii)

Use : .

1. Use suitable identities to find the following products: (iii)

by difference of squares.

1. Use suitable identities to find the following products: (iv)

similarly.

1. Use suitable identities to find the following products: (v)

This is with : result .

1. Use suitable identities to find the following products: (vi)

1. Use suitable identities to find the following products: (vii)

Use : .

1. Use suitable identities to find the following products: (viii)

Using : .

1. Use suitable identities to find the following products: (ix)

2. Find the values using suitable identities: (i)

. Or times directly .

2. Find the values using suitable identities: (ii)

2. Find the values using suitable identities: (iii)

2. Find the values using suitable identities: (iv)

Cannot compute fully easily; use .

2. Find the values using suitable identities: (v)

2. Find the values using suitable identities: (vi)

2. Find the values using suitable identities: (vii)

. Since .

2. Find the values using suitable identities: (viii)

. . So result .

3. Factor the following algebraic expressions: (i)

This is a perfect square: with and , , and . So .

3. Factor the following algebraic expressions: (ii)

This is a difference of squares with and , since . Therefore .

3. Factor the following algebraic expressions: (iii)

This is a difference of cubes with and , since and . Therefore .

3. Factor the following algebraic expressions: (iv)

Need numbers sum product : and . So .

3. Factor the following algebraic expressions: (v)

Group as . So factor is .

3. Factor the following algebraic expressions: (vi)

Sum of cubes: . So .

3. Factor the following algebraic expressions: (vii)

Write it as and apply with . This gives .

3. Factor the following algebraic expressions: (viii)

. So .

3. Factor the following algebraic expressions: (ix)

Multiply through by 3: . Using with : . Dividing back by 3, the expression .

3. Factor the following algebraic expressions: (x)

This is . So factor is .

3. Factor the following algebraic expressions: (xi)

This is a perfect cube with and : , , , . So .

4. Simplify the following: (i) Note: Assume that the denominators are not equal to 0.

Numerator . Denominator . Cancelling (non-zero) gives .

4. Simplify the following: (ii)

Numerator . Denominator . Cancelling : .

4. Simplify the following: (iii)

Numerator . Denominator . Cancelling : .

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units. (i)

. So rectangle could be a square of side units; length , breadth (or any factor pair of the square).

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units. (ii)

Difference of squares: . So length , breadth (assuming positive).

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (i)

Factor: . So dimensions could be or etc. One convenient set: length , breadth , height .

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (ii)

. So dimensions could be .

7. The village playground is shaped as a square of side 40 metres. A path of width metres is created around the playground for people to walk. Find an expression for the area of the path in terms of .

Outer square side . Outer area . Inner playground area . Path area square metres.

8. If a number plus its reciprocal equals , find the number.

Let number be . Then . Multiply: => => . So or .

9. A rectangular pool has area square hastas. If its width is hastas, find its length. Hasta was a unit used to measure length.

Factor area: . Since width is , length is hastas.

10. If both and are factors of , show that .

If is zero: => . If is zero: => => . Equate: => => .

11. If and , then prove that .

We know . Also => => . Then . Multiply by gives . Hence proved.

12. By factoring the expression, check that is always divisible by 6 for all natural numbers . Give reasons.

. This is product of three consecutive natural numbers. Among any three consecutive integers one is divisible by 3 and at least one is even, so product divisible by . Hence .

13. Find the value of (i) , when

Write . Since , , whole expression is . But given expression is , so value is .

13. Find the value of (ii) , when

Note . Write expression as . This is of form with . Since , the identity gives value . Hence .