Factorise by splitting the middle term:-
x^2-120xy-625y^2
To factorize the expression x^2 - 120xy - 625y^2 by splitting the middle term, we need to find two numbers that multiply to give -625y^2 and add up to -120xy.
Let's break down -625y^2 into its prime factors: -1 * 5^4 * y^2.
Now, we need to look for two numbers whose product is -1 * 5^4 * y^2 and whose sum is -120xy. After considering different combinations, we can determine that the numbers are -25y and 25y.
So, we can rewrite the expression as follows:
x^2 - 120xy - 625y^2
= x^2 - 25xy - 95xy - 625y^2
= (x^2 - 25xy) - (95xy + 625y^2)
= x(x - 25y) - 625y(x + 25y)
Thus, the factored form of x^2 - 120xy - 625y^2 is (x - 25y)(x + 25y).
To factorize the given expression using the method of splitting the middle term, we need to find two numbers whose product is equal to the product of the coefficient of the squared term (1) and the constant term (-625) and whose sum is equal to the coefficient of the linear term (-120xy).
The product of the squared term and the constant term is 1 * (-625) = -625.
Let's find the factors of -625: ±1, ±5, ±25, ±125, ±625.
Now, we need to find two numbers whose sum is -120. We can try different combinations of factors to find the correct pair.
The pair that satisfies both conditions is -25 and 5 because (-25) * (5) = -125 and (-25) + 5 = -20.
So we can rewrite the middle term (-120xy) as (-25xy + 5xy):
x^2 - 25xy + 5xy - 625y^2
Now, we group the terms:
(x^2 - 25xy) + (5xy - 625y^2)
Now we can factor out the common terms from each group:
x(x - 25y) + 5y(x - 25y)
Now, notice that we have a common binomial factor (x - 25y) in both terms. We can factor it out:
(x + 5y)(x - 25y)
Therefore, the given expression x^2 - 120xy - 625y^2 can be factorized as (x + 5y)(x - 25y) using the method of splitting the middle term.
625 factors into 5 * 5 * 5 * 5. You have to combine the above factors, so the difference will equal 120.
5 * 5 * 5 = 125
125 - 5 = 120, or even better 5 - 125 = -120
Therefore x^2-120xy-625y^2 = (x+5y)(x-125y)
Although this may not be the process you were seeking, I hope this helps. Thanks for asking.