In Maths, Algebra is one of the important branches. The concept of algebra is used to find the unknown variables or unknown quantity. The multiplication of algebraic expressions is a method of multiplying two given expressions consisting of variables and constants. Algebraic expression is an expression that is built by the combination of integer constants and variables. For example, 4xy + 9, in this expression x and y are variables whereas 4 and 9 are constants. The value of an algebraic expression changes according to the value chosen for the variables of the expressions.

If there are brackets given in any expression, then it should be simplified first. When there is no bracket present, then the algebraic expressions can also be solved by applying division and multiplication, and then addition and subtraction, similar to BODMAS rule.

For a clear idea on this let us take an expression 2x+1. Now if x = 1, the value of expression would be 3. If x = 2 the value will be 5 and so on. The value of the expression is dependent on the value of the variable. And if we have 2(x+1), then we will not get the same answer as we have got earlier. Here, if we put x =1, then it gives 2(1+1) = 4. So, we can see the difference when we use brackets in expressions.

Terms Used in Algebra

Variable: The unknown quantity used in algebraic expressions, such as x, y, z, a, b, etc.

Coefficient: The value attached with the variable is the coefficient of it. For example, for 3x, the coefficient is 3

Monomial: An expression with one term. For example, 4x, 5y, 6z

Binomial: An expression with two terms. For example, 4x+4, 5y+y, 6z+z

Trinomial: An expression with three terms. For example, 4x²+4x+1

Solved Examples

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

Illustration 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.

Illustration 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

Illustration 3: Multiply (2a² + 9a + 10) by 4a.

Solution:

4a × (2a² + 9a + 10)

= (4a × 2a²) + (4a × 9a) + (4a × 10)

= 8a³ + 36a² + 40a

Illustration 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

Illustration 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.

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