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Chapter 11: Algebraic Expressions

    Introduction

    Like and Unlike Terms

    Addition and Subtraction of Algebraic Expressions

    Multiplication of Algebraic Expressions

    Multiplication of a Monomial by a Monomial

    Multiplication of a Binomial or Trinomial by a Monomial

    Multiplication of a Binomial by a Binomial or Trinomial

    What is an Identity ?

    Some Important Identities

    Applications of Identities

    Geometric Verification of the Identities

    Exercise 11.1

    Exercise 11.2

    Exercise 11.3

    Exercise 11.4

    Exercise 11.5

    What We Have Discussed?

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

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Chapter 11: Algebraic Expressions > Multiplication of a Binomial or Trinomial by a Monomial

Multiplication of a Binomial or Trinomial by a Monomial

Expression that contains two terms is called a binomial. An expression containing three terms is a trinomial and so on.

In general, an expression containing, one or more terms with non-zero coefficient (with variables having non negative integers as exponents) is called a polynomial.

Multiplying a monomial by a binomial

Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ?

Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law, 3x × (5y + 2) = (3x × 5y) + (3x × 2) = +

We commonly use distributive law in our calculations.

Instructions

  • 7 × 106 =
  • (Here, we used distributive law)expand the expression is 7 x +
  • Divide the expression with 7 is
  • Multiply the expression with 7 is +
  • we get the answer is =
  • 7 × 38 =
  • (Here, we used distributive law)expand the expression is 7 x
  • Divide the expression with 7 gives us: -
  • we get the answer is =
  • We have found the answer.

We may use commutative law as : 7 × 3 = ×

(or) in general: a × b = b × a

Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = –

And 5xy × y2+3 = 5xy×y2 + 5xy×3 = + .

What about a binomial × monomial?

For example, (5y + 2) × 3x = +

We see that: (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.

Try these

Find the product (i) 2x (3x + 5xy) (ii)a22ab−5c

Instructions

(i) 2x (3x + 5xy) = +
Hence, we get 6x2 + 10x2y
(ii)a22ab−5c = +
Hence, we get 2a3b - 5a2c

Multiplying a monomial by a trinomial

Instructions

Consider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law.
Opening up the brackets, we get: 3p × (4p2 + 5p + 7) = (3p × 4p2) + (3p × 5p) + (3p × 7)
Now, individually multiply all the terms: + +
The answer is 12p3+15p2+12p.

Observe, by using the distributive law, we are able to carry out the multiplication term by term.

Find the product: 4p2+5p+7×3p = 4p2×3p + 5p×3p + 7×3p = + +

Example 5: Simplify the expressions and evaluate them as directed:

(i) x (x – 3) + 2 for x = 1,

Instructions

xx−3+2

  • Expand the bracket - + 2.
  • Substitute value of x is - x + 2.
  • Find the value of each separate constant term
  • Add/subtract them
  • We find the value of expression for x = 1 to be equal to .
  • Found the answer.

(ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2

Instructions

3y2y−7−3y−4−63

  • Expanding 3y (2y – 7) – 3 (y – 4) – 63 we get
  • add the terms 6y2 - y -
  • For apply the y = –2 value in the expression,
  • multiply the values 6 x + 24 x - 51
  • multiply the values here we get + -
  • add the positive terms - 51
  • Hence it is

Example 6: Add (i) 5m3−m and 6m2−13m (ii) 4y3y2+5y−7 and 2y3−4y2+5

(i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = +

Now adding the second expression to it: 15m−5m2 + 6m2 – 13m = +

(ii) The first expression = 4y3y2+5y−7 = 4y×3y2 + (4y × 5y) + (4y × (–7)) = + +

The second expression = 2y3−4y2+5 = 2y3+2×−4y2+2×5 = – +

Thus, 4y3y2+5y−7 + 2y3−4y2+5 = 12y3+20y2−28y + 2y3−8y2+10 = + + +

Example 7: Subtract 3pq (p – q) from 2pq (p + q)

We have 3pqp−q = – and 2pqp+q = +

Thus, 2pq (p + q) - 3pq (p – q) = 2p2q + 2pq2 - (3p2q – 3pq2) = p2q + pq2