what number do you add in the equation x2-16x=23
Half of the coefficient of x, squared. For example: x2 + 16x -3 you would add 8^2. Then you would have: x2 +16x +8^2 = 3-8^2 Which leads to: (x+8)^2 = -61
(the shape is an upside down 'u').
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
(x+8)^2=105
x2+16x-17 = (x-1)(x+17) when factored
In the absence of signs I have assumed the term in 16x and 60 are negative: ie x2 - 16x = -60 or x2 - 16x + 60 = 0 Factors of 60 which total -16 are -6 and -10, so (x - 6)(x - 10)
what number do you add in the equation x2-16x=23
x2+16x-80 = 0 (x-4)(x+20) = 0 x = 4 or x = -20
x2 - 16x + 28 = x2 -2x - 14x - 28 = x(x - 2) - 14(x - 2) = (x - 14)(x - 2)
x2 + 48x + 320
Assuming you wish to make this equation equal zero, for x2 + 16x + 55 = 0, x = -5.
x2-16x+48 = (x-4)(x-12) when factored
12
Half of the coefficient of x, squared. For example: x2 + 16x -3 you would add 8^2. Then you would have: x2 +16x +8^2 = 3-8^2 Which leads to: (x+8)^2 = -61
to factor (x2-16x+64), you need to find two number whose product is 64 and whose sum is 16. These two numbers are 8 and 8. (x2-16x+64)=(x-8)(x-8)
(x - 15)(x - 1)