# NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.3 in Chapter 14 - Factorisation

Question 4 Exercise 14.3

Q2) Divide the given polynomial by the given monomial.

(i) \left(5x^2-6x\right)\div3x

(ii) 3y^8-4y^6+5y^4\div y^4

(iii) 8\left(x^3y^2z^2+x^2y^3z^2+x^2y^2z^3\right)\ \div\ 4y^2x^2z^2

(iv) \left(x^3+2x^2+3x\right)\div\ 2x

(v) \left(p^3q^6-p^6q^3\right)\div\ p^3q^3

Solution 2:

(i) 5x^2-6x\ =\ x\left(5x-6\right)

\frac{5x^2-6x}{3x}\ =\ \frac{x\left(5x-6\right)}{3x}

=\frac{\ 5x-6}{3}

(ii) 3y^8-4y^6+5y^4\ =\ y^4\left(3y^4-4y^2+5\right)

\frac{3y^8-4y^6+5y^4}{y^4}

\frac{y^4\left(3y^4-4y^2+5\right)}{y^4}

3y^4-4y^2+5

(iii) 8\left(x^3y^2z^2+x^2y^3z^2+x^2y^2z^3\right)\ =\ 8x^2y^2z^2\left(x+y+z\right)

Therefore, \frac{8x^2y^2z^2\left(x+y+z\right)}{4x^2y^2z^2}

=2(x+y+z)

(iv) x^3+2x^2+3x\ =\ x\left(x^2+2x+3\right)

\frac{x\left(x^2+2x+3\right)}{2x}

\frac{x^2+2x+3}{2}

(v) p^3q^6-p^6q^3=\ p^3q^3\left(q^3-p^3\right)

Therefore, \frac{p^3q^3\left(q^3-p^3\right)}{p^3q^3}

=\ q^3-p^3

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