Question

42 (a _4 - 13 a_3 + 36a _2 )by 7a (a-4 )

Answer 1

Hey friend, Harish here.

Here is your answer:

To find:

$\frac{42(a ^{4} - 13 a^{3} + 36a ^{2}) }{7a(a-4)}$

Solution:

Now in the numerator take a² as common.

Then,

$\frac{42a^{2}(a^{2} - 13 a + 36)}{7a(a-4)}$

$\frac{42a^{2}}{7a} \times ( \frac{(a^{2} - 13 a + 36)}{(a-4)})$


Now , We know that, 42/7 = 6. and a² & a get canceled. 

Then,

$6a \times ( \frac{(a^{2} - 13 a + 36)}{(a-4)})$

Now factorize the numerator using splitting the middle term:

Then,

⇒ $6a \times( \frac{(a^{2} - 13 a + 36)}{(a-4)}) = 6a\times ( \frac{((a^{2} - 4 a -9a+ 36))}{(a-4)})$

⇒ $6a \times ( \frac{a(a - 4)-9(a-4))}{(a-4)}=6a\times \frac{(a-4)(a-9)}{(a-4)}$

Now, We can cancel (a-4) & (a-4).

Then,

⇒ $6a\times \frac{(a-4)(a-9)}{(a-4)} = 6a \times (a-9)$

⇒ $6a^{2} - 54a$

Therefore the answer is 6a² - 54a.
________________________________________________

Hope my answer is helpful to you.