# Multiplication of Algebraic Expressions

An Algebraic expression is an expression that is built by the combination of integer constants and variables. They undergo operations such as addition, subtraction, multiplication and division. For example: $4xy~ +~ 9$ = $0$, in this expression $x$ and $y$ are variables whereas $4$ and $9$ are constants. The value of algebraic expression changes according to the value chosen for the variables of the expressions.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.

##### Algebraic Expression – Multiplication

Multiplication is simply repeated addition. We multiply variables and constants in an algebraic expression. For example, the area of a rectangular room is the product of length and breadth. The value of area depends on the value chosen for length and breadth. Similarly volume is the product of length, breadth and height.

## Multiplication of Monomial by Monomial

Illustration 1: Multiply $5x$ with $21y$ and $32z$.

Solution: $5x~\times~21y~\times~32z$ = $105xy~\times~32z$ = $3360xyz$.

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: $Volume$ = $length~\times~breadth~\times~height$

Therefore, $volume$ = $5ax~\times~3by~\times~10cz$ = $5~\times~3~\times~10~\times~(ax)~\times~(by)~\times~(cz)$ = $150 axbycz$.

## Multiplying a Monomials and Polynomials

$4a~\times~(2a^2~+~9a~+~10)$ = $(4a~\times~2a^2)~+~(4a~\times~9a)~+~(4a~\times~10)$

= $8a^3~+~36a^2~+~40a$

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

$x(x~-~2)~+~5$

Solution: $x(x~-~2)~+~5$, $x$ = $3$.

Substituting the value of $x$ = $3$

$3~\times~(3~-~2)~+~5$ = $3(1)~+~5$ = $8$.

Illustration 4: 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 $y$ = $-1$.

$4~\times~-1((2~\times~-1)~–~6)~–~3(-1~-~2)~+~20$

= $-4~(-2~-~6)~-~3(-3)~+~20$

= $32~+~9~+~20$ = $61$..