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Last modified on January 10th, 2024

To divide polynomials by monomials, we divide each term of the polynomial in the numerator (dividend) by the monomials in the denominator (divisor) by applying the quotient rule of exponents (a^{m} ÷ a^{n} = a^{m-n}), when needed.

Let us divide the polynomial 5p^{2} – 14p by the monomial p.

Here, we will obtain the quotient by dividing a univariate polynomial by a univariate monomial.

On dividing the terms 5p^{2} and 14p by p, we get

${\dfrac{5p^{2}}{p}=5p^{2-1}=5p}$ and ${\dfrac{-14p}{p}=-14}$

Thus, the quotient of ${\dfrac{5p^{2}-14p}{p}}$ is ${5p-14}$.

Now, let us divide a multivariate polynomial by a univariate monomial.

On dividing the polynomial 10p^{2}q^{3} + 12pq^{4} by the monomial 2q^{2}, we get

${\dfrac{10p^{2}q^{3}+12pq^{4}}{2q^{2}}}$

Dividing the terms 10p^{2}q^{3} and 12pq^{4} by 2q^{2}, yields

${\dfrac{10p^{2}q^{3}}{2q^{2}}=\dfrac{10}{2}p^{2}q^{3-2}=5p^{2}q}$

${\dfrac{12pq^{4}}{2q^{2}}=\dfrac{12}{2}pq^{4-2}=6pq^{2}}$

Thus, the quotient is 5p^{2}q + 6pq^{2}.

Similarly, we can divide a multivariate polynomial 11xyz – 7xy^{2}z^{3 }by the multivariate monomial xyz.

The quotient is:

${\dfrac{11xyz-7xy^{2}z^{3}}{xyz}}$

= ${\dfrac{11xyz}{xyz}-\dfrac{7xy^{2}z^{3}}{xyz}}$

= ${11-7y^{2-1}z^{3-1}}$

= ${11-7yz^{2}}$

**Simplify: (18m ^{7} + 12m^{6} – 24m^{5}) ÷ 6m^{4}**

Solution:

${\dfrac{18m^{7}+12m^{6}-24m^{5}}{6m^{4}}}$

= ${\dfrac{18m^{7}+}{6m^{4}}\dfrac{12m^{6}}{6m^{4}}-\dfrac{24m^{5}}{6m^{4}}}$

= ${\dfrac{18}{6}m^{7-4}+\dfrac{12}{6}m^{6-4}-\dfrac{24}{6}m^{5-4}}$

= ${3m^{3}+2m^{2}-4m}$

Thus, (18m^{7} + 12m^{6} – 24m^{5}) ÷ 6m^{4} is simplified to 3m^{3} + 2m^{2} – 4m.

Another method for dividing the polynomials by monomials involves finding their common factors and canceling them.

Let us divide 4u^{7}v^{3} – 6v^{7}u^{5} by 2u^{5}v^{3} using this method.

By finding their common factors, we get

${\dfrac{4u^{7}v^{3}-6v^{7}u^{5}}{2u^{5}v^{3}}}$

= ${\dfrac{2u^{5}v^{3}\left( 2u^{2}-6v^{4}\right) }{2u^{5}v^{3}}}$

Now, by canceling out the common factors from the numerator and the denominator, we get

${2u^{2}-6v^{4}}$

Thus, the quotient is 2u^{2} – 6v^{4}.

**Divide: (-40a ^{8} – 32a^{7} + 88a^{11} + 16a^{2}) ÷ 8a^{2}**

Solution:

${\dfrac{-40a^{8}-32a^{7}+88a^{11}+16a^{2}}{8a^{2}}}$

= ${\dfrac{8a^{2}\left( -5a^{6}-4a^{5}+11a^{9}+2\right) }{8a^{2}}}$

= ${-5a^{6}-4a^{5}+11a^{9}+2}$

Thus, the quotient is 11a^{9} – 5a^{6} – 4a^{5} + 2.

Last modified on January 10th, 2024