Question

factorise a^2-10ab-75b^2​

Answer 1

Answer:

a^2 - 10 ab - 75 b^2

a^2 - ( 15 - 5 ) ab - 75 b^2

a^2 - 15 ab + 5 ab - 75 b^2

a( a - 15 b ) + 5b ( a - 15 b )

(a - 15 b) + ( a + 5b )

Answer 2

Factorization of  $a^{2} -10ab -75b^{2}$  is (a+5b)(a-15b).

Following are the steps to be followed to factorize any given 2-degree$\alpha , \beta , and c$ equation:

1. For any equation of the form $\alpha a^{2}+ cab+\beta b^{2}$ where $\alpha , \beta$ and c are constants, firstly make sure that the coefficient of the term $a^{2}$ is equal to 1.
2. If it is not, make it 1 by dividing the complete equation by that coefficient (here it is $\alpha$).
3. Now, the equation will look like this $a^{2} + \frac{c}{\alpha } ab + \frac{\beta }{\alpha }b^{2}$ . However, in our given question the coefficient of the term $a^{2}$ is already 1.
4. Now, we would have to find the factors of the coefficient of the term $b^{2}$ such that the addition or subtraction of those two terms is equal to the coefficient of the middle term.

Let us apply this method here,

Our given 2-degree equation is:  $a^{2} -10ab -75b^{2}$  

⇒ Here, the coefficient of the term $b^{2}$ is -75. So factors of -75 are as follows.

⇒ -75 = -1 × 3 × 5 × 5.

⇒ Now, we want such a combination where the difference between two terms is equal to -10ab (i.e., the middle term)

⇒ One such combination is '-15' and '5' as its multiplication gives us the  coefficient of $b^{2}$ and their sum gives us the middle term.

Now, we will split the middle term of our equation into two using the above combination and will simplify further to get the factors;

∴ $a^{2} -10ab -75b^{2}$  =  $a^{2} -15ab +5ab -75b^{2}$

                               = $(a^{2}-15ab )+(5ab -75b^{2} )$

                               = $a(a-15b)+5b(a-15b)$

∴  $a^{2} -10ab -75b^{2}$  = (a+5b)(a-15b).

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