Question

Разложите на множители 1)81-(x^2+6x)^2, 1)c^2+4c+4-k^2, 3)a^2-b^2-10b-25.

Answer 1

1) 81-(x^2+6x)^2=(9-x^2-6x)(9+x^2+6x)
2) c^2+4c+4-k^2=(c+2)^2-k^2=(c+2-k)(c+2+k)
3) a^2-b^2-10b-25=a^2-(b^2+10b+25)=a^2-(b+5)^2=(a-b-5)(a+b+5)

Answer 2

Разложение на множители

1) 81 - (x^2 + 6x)^2

To factorize this expression, let's first expand the square term:

81 - (x^2 + 6x)^2 = 81 - (x^2 + 2*6x + 6^2) = 81 - (x^2 + 12x + 36)

Now, we have a difference of squares:

81 - (x^2 + 12x + 36) = 9^2 - (x + 6)^2

Using the formula for the difference of squares, we can rewrite this expression as:

(9 - (x + 6))(9 + (x + 6)) = (9 - x - 6)(9 + x + 6) = (3 - x)(15 + x)

So, the expression 81 - (x^2 + 6x)^2 can be factored as (3 - x)(15 + x).

2) c^2 + 4c + 4 - k^2

This expression doesn't seem to be factorable using simple methods. However, we can simplify it by noticing that the first three terms form a perfect square trinomial:

c^2 + 4c + 4 = (c + 2)^2

So, the expression can be rewritten as:

(c + 2)^2 - k^2

This is a difference of squares:

(c + 2 + k)(c + 2 - k)

Therefore, the expression c^2 + 4c + 4 - k^2 can be factored as (c + 2 + k)(c + 2 - k).

3) a^2 - b^2 - 10b - 25

This expression can be factorized using the difference of squares formula for the first two terms:

a^2 - b^2 = (a + b)(a - b)

Now, let's look at the last two terms:

-10b - 25 = -5(2b + 5)

Putting it all together, the expression a^2 - b^2 - 10b - 25 can be factored as (a + b)(a - b) - 5(2b + 5).

I hope this helps! Let me know if you have any further questions.