Product of monomial by a monomial — lesson. Mathematics CBSE, Class 8.

PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

Book Free Demo

Multiplying two monomial:

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 monomials.

Example:

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

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

$-2x^2, 4xy^3$ 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}

Previous theory

Exit to the topic

Next theory

Send feedback

Did you find an error?

Send it to us!

4. Product of monomial by a monomial

Theory: