Assuming multiplication, it's 4a2.
4a3 - 4a2 = 4a2*(a - 1)
4a2+ 25 does not factor over the real number field. In the complex numbers , it factors as (2a +5i)(2a - 5i). This is because i2 = -1, so 4a2 + 25 = 4a2 - (- 25) = 4a2 - 25(-1) = 4a2 - 25i2
If a is an integer then 4 divides 4a2 so it cannot be a prime. If a = 0 then 4a2 = 0 and cannot be simplified further If a = 1 then 4a2 = 4 = 2*2 If a is an integer > 1, then 4a2 = 2*2*a*a If a is not an integer, then there are too many possible options for the answer.
(2a + 5)(2a - 5)
4a2 - 20ab2 + 25b4 (2a - 5b2)(2a - 5b2) (2a - 5b2)2
It is 4a2.
(2a + 1)(2a - 5)
4a2 + 4ab - y2 + b2 You cannot simplify this expression because it does not contain like terms, but you can factor it such as: 4a2 + 4ab - y2 + b2 = (4a2 + 4ab + b2) - y2 = (2a + b)2 - y2 = [(2a + b) - y][(2a + b) + y]
2(2a2 - 31)
Assuming the missing signs are pluses, that factors to (4a + 3)(a + 3)
The LCM of 4a2b and 6a10b is 12a10b
(4a - 1)(a - 4)
4a2 - 9a + 5 = 4a2 - 4a -5a + 5 = 4a(a - 1) - 5(a - 1) = (4a - 5)(a - 1)
2a2+2a2 = 4a2
2(a - 1)(2a + 1)
= 4a2 + 2ab 2b2
a = [ 0, 4 ]
(2a + 10)(2a - 12)
GCF(28a5, 16a2) = 4a2.
The essence of the proof is simply to complete the square for a generalised quadratic equation. Like this:ax2 + bx + c = 0Take 'a' outside:a[x2 + bx/a + c/a] = 0Divide through by 'a':x2 + bx/a + c/a = 0Complete the square:(x + b/2a)2 - b2/4a2 + c/a = 0Rearrange to find x:(x + b/2a)2 = b2/4a2 - c/ax + b/2a = (+/-)sqrt[b2/4a2 - c/a]x = -b/2a (+/-) sqrt[b2/4a2 - c/a]Finally, fiddle around so that (1/2a) can be taken out as a common factor:x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.
(x + 4)(4x - 1)