To factor this equation... if you have learned before, find 2 numbers that add up to the middle value (-7) and multiply together to have a product of the last value (+10).
You can use guess and check for this problem... but if you haven't already found it... the sum of -5 and -2 is -7... and the product of -5 and -2 is +10, so we can use the values -5 and -2....
The solution is (c-5)(c-2) = c^2 - 7c + 10.
k = 7 x^4 - 5x^3 + 7x^2 + 3x - 10 = (x + 1)(x - 2)(x^2 - 4x + 5)
x2 + 100 has no real solutions. Applying the quadratic formula, we find two imaginary solutions: Zero plus or minus 10itimes the square root of 1.x = 10ix = -10iwhere i is the square root of negative 1.
The correct factorization of 5 x plus 2 -3x x plus 2 is 2x+4.
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: 10 plus or minus 4 times the square root of 7. x = 20.583005244258363 x = -0.5830052442583629
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (-3 plus or minus the square root of 12329) divided by 2. x = 54.01801509420163 x = -57.01801509420163
(5x - 2)(x - 5)
20x2 - 3x + 10 does not have any real factors.
x2+2x-120 = (x+12)(x-10) when factored
The factors are [3x + 5 - 2*sqrt(10)] and [3x 5 + 2*sqrt(10)].
2
n2 - 100 = (n + 10)(n - 10)
10x^2 - x - 2 = (5x + 2)(2x - 1)
(3x-2)(x-2)
(x + 5)(x + 2)
That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: (-5 plus or minus the square root of -15) divided by 2.x = -2.5 + 1.9364916731037085ix = -2.5 - 1.9364916731037085iwhere i is the square root of negative one.
(3r + 2)(r - 5)
(11x - 2)(3x - 5)