A) x=-2,1,3(x+2)(x-1)(x-3)=0(x+2)(x2-4x+3)=0x3-4x2+3x+2x2-8x+6=0x3-2x2-5x+6=0 b) x=-3,3,i and also -i(x-3)(x+3)(x-i)(x+i)=0(x2-9)(x2-i2)=0(x2-9)(x2+1)=0x4-9x2+x2-9=0x4-8x2-9=0 c) x=-2,-2,2-3i (also 2+3i), 4+√2 (also 4-√2)(x+2)(x+2)(x-[2-3i])(x-[2+3i])(x-[4+√2])(x-[4-√2])(x2+4x+4)(x2-[2-3i]x-[2+3i]x+4-9i2)(x2-[4+√2]x-[4-√2]x+16-2)(x2+4x+4)(x2-4x+13)(x2-8x+14)you now multiply all three trinomials together
Answer:
6m^2-17m+12
Step-by-step explanation:
Answer:
x = 6/5
Step-by-step explanation:
3x - 3/5 + 2 = 5
<u>Add 3/5 and -2 on both sides.</u>
3x = 5 + 3/5 - 2
3x = 18/5
<u>Divide 3 on both sides.</u>
x = 18/5 ÷ 3
x = 18/15
x = 6/5
Answer:
13 can go into 52.
Step-by-step explanation:
52/13=4
something noteworthy is that the independent and squared variable in this case will be the "x", namely the graph of that quadratic is a vertical parabola.
![\bf f(x) = (x+2)(x-4)\implies 0=(x+2)(x-4)\implies x = \begin{cases} -2\\ 4 \end{cases} \\\\\\ \boxed{-2}\rule[0.35em]{7em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}\stackrel{\downarrow }{1}\rule[0.35em]{10em}{0.25pt}\boxed{4}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%28x%2B2%29%28x-4%29%5Cimplies%200%3D%28x%2B2%29%28x-4%29%5Cimplies%20x%20%3D%20%5Cbegin%7Bcases%7D%20-2%5C%5C%204%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-2%7D%5Crule%5B0.35em%5D%7B7em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B3em%7D%7B0.25pt%7D%5Cstackrel%7B%5Cdownarrow%20%7D%7B1%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%5Cboxed%7B4%7D)
so the parabola has solutions at x = -2 and x = 4, and its vertex will be half-way between those two guys, namely at x = 1.
since this is a vertical parabola, its axis of symmetry, the line that splits its into twin sides, will be a vertical line, and it'll be the x-coordinate of the vertex, since the vertex hasa a coordinate of x = 1, then the axis of symmetry is the vertical line of x = 1.
