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Chapter 8: Algebraic Expressions and Identities

    Introduction

    Addition and Subtraction of Algebraic Expressions

    Multiplication of Algebraic Expressions: Introduction

    Multiplying a Monomial by a Monomial

    Multiplying a Monomial by a Polynomial

    Multiplying a Polynomial by a Polynomial

    Exercise 8.1

    Exercise 8.2

    Exercise 8.3

    Exercise 8.4

    What Have We Discussed ?

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Chapter 8: Algebraic Expressions and Identities > Multiplying a Monomial by a Polynomial

Multiplying a Monomial by a Polynomial

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

  • Finding 7 × 106
  • (Here, we used distributive law)expand the expression is 7 x ( + )
  • Taking 7 as a common factor.
  • Multiplying the expression: +
  • We get the answer is
  • 7 × 38 =
  • (Here, we used distributive law)Expand the expression: 7 x
  • Multiplying 7 through out gives us: -
  • We get the answer as
  • 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