can these problems be factored??

Question

x^2+x+6??????


or


4k^2+20k+25????



then last




4x^2-24x+36

Answer 1

The first one cannot be factored, but the second two can:

1) No factored version

2) 4k^2+20k+25 = (5+2k)^2

3) 4x^2-24x+36 = 4(3-x)^2

I hope this helps!

Answer 2

First one, no. x² + x + 6 is not factorable.

4k² + 20k + 25 = (2k + 5)²

4x² - 24x + 36 = 4(x² - 6x + 9) = 4(x - 3)²

Answer 3

1) No factored version

2) 4k^2+20k+25 = (5+2k)^2

3) 4x^2-24x+36 = 4(3-x)^2

Answer 4

Assume we are factoring over the integers.
If you have a quadratic ax² + bx + c, b² -4ac must
be a square for such a factorisation to exist.
So let's tackle your questions 1 at a time.
a). x² + x + 6. Here b² -4ac = 1- 24 = -23, which is not a square.
No.

b. 4k^2 +20k + 25. Now b² -4ac = 0, so we can factor this.
In fact the factorisation is (2k+5)².

c. First take out the common factor of 4 to get
4(x² -6x + 9). Now go to work on the term in brackets.
Again, b² - 4ac = 0 and the factorisation is 4(x-3)².
As you might have guessed, b² - 4ac = 0 means
the quadratic is a square trinomial.

Answer 5

x^2+x+6
(x+3)(x-2)

4k^2+20k+25
(2k+5)^2

4x^2-24x+36
(2x-6)^2

Answer 6

1.(x+3)(x+2)

2.2k+5)^2

3.(2x-12)^2 or 4(x-3)^2