Multiplication of monomial with polynomial terms — lesson. Maths CBSE Live product, Class 8.

Multiplying two monomial:

A monomial is a polynomial with exacty $1$ term.

Let's take two monomials $6$ and $y$.

We can multiply and write them as $6 × y = 6y = y +y +y +y +y + y$ (or) adding $y$ for $6$ times.

This is easy to multiply because one monomial is a variable and one monomial is a number.

If we have $2$ monomials with variables and numbers, then the product is the result of (product of coefficients of monomials) $×$ (product of variables of monomials).

Example:

To multiply $4x$ and $5y$.

The coefficients are $4$ and $5$, and the variables are $x$ and $y$.

$4x× 5y =$ ($4×5$)$×$($x×y$) $= 20xy$.

To multiply monomials of higher degrees, find the product of coefficients of monomials and then use the law of exponents to add the power of similar variables.

The useful law of exponents are as follows:

$a^{m} \times a^{n} = a^{m + n}$
${(a^{m})}^{n} = a^{mn}$

If we have both the monomials as the same variables, then use the law of exponents,

a^{m} \times a^{n} = a^{m + n}

to find the product. For example,

y^{2} \times y^{2} = y^{2 + 2} = y^{4}

. This is not feasible in case of different variables.

Example:

If we have to multiply monomials of higher degrees, such as

4 x^{3} and 5 x y^{4}

, it will be

20 x^{4} y^{4}

, find the product of coefficients of monomials $= 20$ and then use the law of exponents to add the power of similar variables,

a^{m} \times a^{n} = a^{m + n}

to find

x^{3} \times x^{1} = x^{4}

and then

y^{4}

. Hence, we have

20 x^{4} y^{4}

Multiplying more than two monomials:

To multiply more than two monomials, extend the idea of multiplying two monomials to more than two monomials.

This can be understood with the help of an example having three monomials.

Example:

Find the volume of a cuboid whose sides are $-2x^2, 4xy^3$ and $-5y^2z^3$.

$-2x^2, 4xy^3$ and $-5y^2z^3$

Let us multiply the three monomials.

$= (-2×4×-5) × (x^2×xy^3×y^2z^3)$

$= (-×+×-) 2×4×5$ $((x^2×x)×(y^3×y^2)×(z^3))$

= 40 (x^{3} \times y^{5} \times z^{3})

= 40 x^{3} y^{5} z^{3}

Multiplying a monmial with a binomial:

A binomial is a polynomial with exactly $2$ terms.

Let us recall the distributive property.

If $a$ is a constant, $x$ and $y$ are variables, then $a(x + y) = ax + ay$.

The distributive property of multiplication over addition or subtracton is indeed a product of a monomial with a binomial.

Let us consider an example.

Example:

Find the product of $3xy$ and $(5x + 9xz)$.

Applying the distributive property, we get:

$3xy$ $\times$ $(5x + 9xz)$ $=$ $(3xy \times 5x) + (3xy \times 9zx)$

$=$ $(3\times 5 x ^{1+1} y) + (3 \times 9 x^{1+1} yz)$

$=$ $15 x^{2} y + 27 x^{2} yz$

Multiplying monomial with a trinomial:

A trinomial is a polynomial with exactly $3$ terms.

Example:

1. Suppose there will be $x$ number of bags and a bag contains $3$ cupcakes of '$p$' packs, $7$ chocolates of '$q$' packs and $5$ cookies of '$r$' packs. The total number of items can be identified by adding the number of items in the bag and product with the number of bags.

This can be written as

x (3 p + 7 q + 5 r)

.

Applying the distributive property,

$= 3px + 7qx + 5rx$.

2. Find the product of

3 p^{3} q

and

(2 {pq}^{3} - 5 p^{2} + 3 q^{4})

= 3 p^{3} q \times (2 {pq}^{3} - 5 p^{2} + 3 q^{4})

Applying the distributive property,

$=$ $(3×2)$ $p^{3+1}q^{1+3}$ $+$ $(3×-5)$$p^{3+2}q$ $+$ $(3×3)p^3q^{1+4}$

$=$ $6p^4q^4-15p^5q+9p^3q^5$.

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1. Multiplication of monomial with polynomial terms

Theory: