8X^2 + 2X - 1 = 0 need quadratic formula -b +/- sqrt(b^2-4ac)/2a a = 8 b = 2 c = -1 -2 +/- sqrt(2^2 -4(8)(-1))/2(8) -2 +/- sqrt(4 + 32)/16 -2 +/- sqrt(36)/16 (-2 +/- 6)/16 X = 1/4 X = -1/2 My TI-84 confirms this answer
(x + 2)(x - 7)
x2(x - 8)
80
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (8 plus or minus the square root of 10) divided by 3 x = 3.720759220056127 x = 1.6125741132772067
8x2 + 2x - 1 = 8x2 + 4x - 2x - 1 = 4x(2x + 1) - 1(2x + 1) = (4x - 1)(2x + 1)
8X^2 + 2X - 1 = 0 need quadratic formula -b +/- sqrt(b^2-4ac)/2a a = 8 b = 2 c = -1 -2 +/- sqrt(2^2 -4(8)(-1))/2(8) -2 +/- sqrt(4 + 32)/16 -2 +/- sqrt(36)/16 (-2 +/- 6)/16 X = 1/4 X = -1/2 My TI-84 confirms this answer
What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).
8x2 - 12x - 8 = 4 (2x2 - 3x - 2) = 4 (2x2 + x - 4x - 2) = 4 (x (2x + 1) - 2 (2x + 1)) = 4 (x - 2) (2x + 1).
Simplifying 8x2 + -8x + -16 Factor out the Greatest Common Factor (GCF), '8'. 8(x2 + -1x + -2) Factor a trinomial. 8((-x + -1)(-x + 2)) Final result: 8(x - 2)(x + 1)
(4x - 1)(2x + 1)
They are: (4x-5)(2x+5)
-8x2 - 2x + 8 this is a quadratic equation or a second order polynomial it is a second order polynomial because it has a term in x2 For every polynomial we name it according to the highest power term in the equation.......
You add compatible terms together. For example, if one equation has 3x2, and the other has 5x2, you combine those two terms as 8x2.
-4x2 - 32x + x + 8 = 8x2 - 56x : simplify and group the various terms-32x + 56x + x + 8 = 8x2 + 4x225x + 8 = 12x2 : Now present this as a standard quadratic equation.12x2 - 25x - 8 = 0 : Solving for the roots of a quadratic equation gives :-x = [25 ± √(252 + 4x12x8)] ÷ 2x12 = [25 ± √1009] ÷ 24 = [25 ± 31.765] ÷ 24x = 56.765/24 = 2.365 and x = -6.765/24 = -0.282x = 2.365 and -0.282
(x + 2)(x - 7)
8x2 + 6x - 9 = 8x2 + 12x - 6x - 9 = 4x(2x + 3) - 3(2x + 3) = (2x + 3)(4x - 3)