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

Question 7 Exercise 14.3

Q4) Divide as directed.

(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4)

(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)

(iv) 20\left(y+4\right)\left(y^2+5y+3\right)\div5\left(y+4\right)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

Solution 4:

(i) \frac{5\left(2x+1\right)\left(3x+5\right)}{2x+1}

= 5(3x+5)

(ii) \frac{2\times13\times x\times y\times\left(x+5\right)\times\left(y-4\right)}{13\times x\times\left(y-4\right)}

=2\times y\times\left(x+5\right)

=2y\left(x+5\right)

(iii) \frac{2\times2\times13\times p\times q\times r\times\left(p+q\right)\times\left(q+r\right)\times\left(r+p\right)}{2\times2\times2\times13\times p\times q\times\left(q+p\right)\times\left(r+p\right)}

=\frac{r\left(p+q\right)}{2}

(iv) \frac{20\left(y+4\right)\left(y^2+5y+3\right)}{5\left(y+4\right)}

=4\left(y^2+5y+3\right)

(v)\frac{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)}

(x+2)(x+3)

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