It is; (2x+4)^2.
True
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
5x - 16 = 3x2x - 16 = 02x = 16x = 8
f(x) = -4x2 - 16x - 11 a = -4, b = -16, c = -11 x-coordinate of the vertex = -b/2a = -(-16)/2(-4) = 16/-8 = -2 y-coordinate = f(-2) = -4(-2)2 -16(-2) - 11 = -16 + 32 - 11 = 5 vertex is (-2, 5)
It is; (2x+4)^2.
4(x + 2)(x + 2)
The Commutative Property of Addition.
b. -4x2+20 = 4x2 = 4-20x2 = -16x = -4Note that -42 = -16 but (-4)2 in brackets = 16
True
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
x+15=16x=16-15x=1
First you need to solve for y. So write 4x2+y=16 so y=16-4x2 Now write f(x)=16-4x2
8x2 + 16x + 8 = 8*(x2 + 2x + 1) = 8*(x + 1)2
x= 2 because 2x+4=16 4x2=8 8+2 or 8x2=16
16x = 4 Divide both sides by '16' Hence 16x/16 = 4/16 x = 4 / 16 x = 1/4
5x - 16 = 3x2x - 16 = 02x = 16x = 8