Multiplication of Algebraic Expressions

Let us solve some problems here based on the multiplication of different types of algebraic expressions.

1: Multiply 5x with 21y and 32z

Solution: 5x × 21y × 32z = 105xy × 32z = 3360xyz

We multiply the first two monomials and then the resulting monomial to the third monomial.

2: Find the volume of a cuboid whose length is 5ax, breadth is 3by and height is 10cz.

Solution:

Volume = length × breadth × height

Therefore, volume = 5ax × 3by × 10cz = 5 × 3 × 10 × (ax) × (by) × (cz) = 150axbycz

3: Multiply (2a² + 9a + 10) by 4a.

Solution:

4a × (2a² + 9a + 10)

= (4a × 2a²) + (4a × 9a) + (4a × 10)

= 8a³ + 36a² + 40a

4: Simplify the below algebraic expression and obtain its value for x = 3.

x(x − 2) + 5

Solution: Given, x(x − 2) + 5, x = 3.

On simplifying the given expression, we get:

x²-2x+5

Now putting x = 3, we get;

= 3²-2(3)+5

= 9 – 6 + 5

= 8

5: Simplify the below algebraic expression and obtain its value for y = −1

4y(2y − 6) – 3(y − 2) + 20

Solution: 4y(2y − 6) − 3(y − 2) + 20 for y = −1

Substituting the value of y = −1.

4 × −1((2 × −1) – 6) – 3(−1 − 2) + 20

= −4 (−2 − 6) − 3(−3) + 20

= 32 + 9 + 20 = 61.

In the division of an algebraic expression, we cancel the common terms, which is similar to the division of numbers. Division of algebraic expressions involves the following steps.

• Step 1: Directly take out common terms or factories in the given expressions to check for the common terms.
• Step 2: Cancel the common term.

Note: Here, the common terms correspond to either of the following: constants, variables, terms, or just coefficients.

There are different types of division of algebraic expressions.

• Division of monomial by a monomial
• Division of polynomial by a monomial
• Division of polynomial by a polynomial

In any case, we first take out common terms from the given polynomials and then eliminate that common term/terms. 

A monomial is a type of expression that has only one term. The correct method to perform the division of a monomial by another monomial is given below:

Consider an example, 27x³÷3x

Here 3x and 27x³ be the two monomials.

• Write their prime factorization. 27x³ ÷ 3x = 27×x×x×x/3×x
• Cancel the common term, which is 3x.

Thus, 27x³ ÷ 3x = 9x²

A polynomial contains a few types of expressions, some of which are binomial, trinomial, or an equation with n-terms.

Now, let’s perform dividing polynomials by monomials.

(4y³ + 5y² + 6y) ÷ 2y

Here, the trinomial is 4y³ + 5y² + 6y, and the monomial is 2y.

• In trinomial, on taking the common factor 2y, it becomes: 4y³ + 5y² + 6y = 2y(2y² + (5/2)y + 3)
• Now, we do the division operation: {2y(2y² + (5/2)y + 3)} ÷ 2y. Cancel 2y from the numerator and the denominator: (4y³ + 5y² + 6y) ÷ 2y = 2y² + (5/2)y + 3

Thus, (4y³ + 5y² + 6y) ÷ 2y = 2y² + (5/2)y + 3

(7x² + 14x) ÷ (x + 2)

Here, both polynomials exist in the binomial form.

• Take out the common factors. For the polynomial 7x² + 14x, x is the common factor.
• So, consider “7x” as a common factor among them. Then it becomes, 7x² + 14x = 7x(x+2)
• Now, (7x² + 14x) ÷ (x + 2) = 7x(x + 2) / (x + 2)
• Eliminate (x+2) from the numerator and denominator, we get the solution for the long dividing polynomials as: (7x² + 14x) ÷ (x + 2) = 7x

Thus, (7x² + 14x) ÷ (x + 2) = 7x