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

Question 10 Exercise 14.3

Q5)Factorise the expressions and divide them as directed.

(i) y^2+7y+10\ \div\ \left(y+5\right)

(ii) \left(m^2-14m-32\right)\div\ m+2

(iii) 5p^2-25p+20\ \div\ p-1

(iv) 4yz\left(z^2+6z-16\right)\div\ 2y\left(z+8\right)

(v) 5pq\left(p^2-q^2\right)\div2p\left(p+q\right)

(vi) 12xy\ \left(9x^2-16y^2\right)\div4xy\left(3x+4y\right)

(vii) 39y^3\left(50y^2-98\right)\div\ 26y^2\left(5y+7\right)

Solution 5:

(i) y^2+7y+10\ =\ y^2+2y+5y+10

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

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

Division: \frac{\left(y+5\right)\left(y+2\right)}{y+5}=\ y+2

(ii) m^2-14m-32\ =\ m^2-16m+2m-32

=\ m\left(m-16\right)+2\left(m-16\right)

=\ \left(m+2\right)\left(m-16\right)

Division: \frac{\left(m+2\right)\left(m-16\right)}{m+2}

= m-16

(iii) 5p^2-25p+20\ =\ 5\left(p^2-5p+4\right)

= 5\left(p^2-4p-p+4\right)

=5\left[p\left(p-4\right)-1\left(p-4\right)\right]

=5\left(p-1\right)\left(p-4\right)

Division: \frac{5\left(p-1\right)\left(p-4\right)}{p-1}

= 5(p-4)

(iv) 4yz\left(z^2+6z-16\right)=4yz\left[z^2+8z-2z-16\right]

=4yz\left[z\left(z+8\right)-2\left(z+8\right)\right]

4yz\left(z-2\right)\left(z+8\right)

division = \frac{4yz\left(z-2\right)\left(z+8\right)}{2y\left(z+8\right)}\ =\ 2z\left(z-2\right)

(v) 5pq\left(p^2-q^2\right)=5pq\left(p+q\right)\left(p-q\right)

Division: \frac{5pq\left(p+q\right)\left(p-q\right)}{2p\left(p+q\right)}

=\ \frac{5}{2}q\left(p-q\right)

(vi) 12xy\left(9x^2-16y^2\right)=\ \ 2\times2\times3\times x\times y\times\left[\left(3x\right)^2-\left(4y\right)^2\right]

=3\times4xy\times\left[\left(3x+4y\right)\left(3x-4y\right)\right]

Division: \frac{3\times4xy\times\left(3x+4y\right)\left(3x-4y\right)}{4xy\left(3x+4y\right)}

= 3(3x-4y)

(vii) 39y^3\left(50y^2-98\right)=\ 3\times13\times y\times y\times y\times2\left(25y^2-49\right)

=3\times13\times y\times y\times y\times2\left(\left(5y\right)^2-\left(7\right)^2\right)

=3\times13\times2\times y\times y\times y\times\left[\left(5y+7\right)\left(5y-7\right)\right]

26y^2\left(5y+7\right)=2\times13\times y\times y\times\left(5y+7\right)

Division = \frac{2\times3\times13\times y\times y\times y\times\left(5y+7\right)\left(5y-7\right)}{2\times13\times y\times y\times\left(5y+7\right)}

=\ 3\times y\times\left(5y-7\right)

=3y(5y-7)

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