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

Factorize the expressions and divide them as directed:

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

Solution:

Factorization is the process of breaking down one object (such as a number, matrix, or polynomial) into a product of another object, or factors, which when multiplied together produce the original object.

\begin{aligned} &\text { First solve for } 50 y^{2}-98, \text { we have }\\ &50 y^{2}-98=2\left(25 y^{2}-49\right)=2\left((5 y)^{2}-7^{2}\right)=2(5 y-7)(5 y+7)\\ &\text { Now, } 39 y^{3}\left(50 y^{2}-98\right) \div 26 y^{2}(5 y+7)=\frac{3 \times 13 \times y^{8} \times 2(5 y-7)(5 y+7)}{2 \times 13 \times y^{2}(5 y+7)}=3 y(5 y-7) \end{aligned}

Video transcript
"hello students i am rita your math leader tutor and the today's question is factorize the expressions and divide them as directed and the question is 39 y raised to power 3 bracket 50 y square minus 98 divided by 26 y raised to the power 2 bracket 5 y plus 7 first of all in this question we take out common 2 from the 50 y square and 98 we take out common 2 from them then we got 25 y square minus 49 divided by 26 y square five y plus seven now 25 is the perfect square of five and 25 is the perfect square of 5 and 49 is a perfect square of 7 so we can write it as 7 square divided by 26 y raised to power 2 5 5 plus 7 now as we know the formula of e square minus b square is equal to a plus b bracket a minus b now we use this identity and solve the equation so 39 into 2 y raised to power 3 into a plus b 5 y plus 7 into 5 5 minus 7 divided by 26 y raised to power 2 bracket 5 y plus 7 now we cancel out the terms 5 y plus seven and thirteen twos are thirteen threes are and two ones are two ones are so we got the answer and the answer is 3 and y is to power 3 and y is to power 2 so 3 minus 2 we got 1 so single y y is 2 power 1 bracket 5 y minus 7 so this is our answer i hope you like this video so please subscribe leo for more updates and do comment your questions thank you"
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