(2x - 3)(2x + 3)
(2x-3)(2x+3)
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x2 - 5x = 0Factor the left side:x (x - 5) = 0x = 0x = 5
x3+13x2=-22xadd 22x to both sides:x3+13x2+22x=0now factor:x(x2+13x+22)=0x(x+11)(x+2)=0x=0x+11=0x+2=0Therefore the solution to this equation is x=0, x=-11, x=-2
Answer : x = ± 1 4x2 - 30 = 34 : 4x2 = 34 - 30 = 4 4x2 = 4 : x2 = 1 therefore x = √1 = ± 1
This is how I did it:8x3+4x2+6x+3[8x3+4x2]+[6x+3][(4x2)(2x+1)]+[(3)(2x+1)]....as you can see, this is in the general form of ab+cb, which factored is b(a+c), so:(2x+1)(4x2+3)
f'(x)= 0x^-1=0anything multiplied by zero equals zero