It is a quadratic polynomial with integer coefficients..
x2 + 3x - 40 = (x + 8)(x - 5)
3x3-18x2+24x=3x(x2-6x+8)=3x(x-4)(x-2)
(x + 8)(x - 5)
Suppose the number is x. Then x2 = 5x + 24 or x2 - 5x - 24 = 0 Then x2 - 8x + 3x - 24 = 0 or x(x - 8) + 3(x - 8) = 0 that is (x - 8)*(x + 3) = 0 so that x = 8 or x = -3 Since the number is a natural number, it cannot be -3, and so the answer is 8.
It is a quadratic polynomial with integer coefficients..
x2 + 3x - 40 = (x + 8)(x - 5)
x2 + 3x - 40 = x2 + 8x - 5x - 40 = x(x + 8) - 5(x + 8) = (x - 5)(x + 8)
3x3 + 6x2 - 24x = 3x(x2 + 2x - 8) = 3x(x2 + 4x - 2x - 8) = 3x[ x(x + 4) - 2(x + 4) ] = 3x(x - 2)(x + 4)
(x + 5)(x - 8)
x2 - 3x - 40 = (x - 8) (x + 5)
x^(2) + 3x - 7 = 5x + 8 Subtract '8' from both sides Hence x^(2) + 3x - 15 = 5x Subtract '5x' from both sides Hence x^(2) -2x - 15 = 0 It is now in quadratic form and will factor. Hence ( x -5)(x + 3) = 0 x = 5 & x = -3
x2+3x+8 is a trinomial But this quadratic equation will have no solutions because its discriminant is less than zero.
3x3-18x2+24x=3x(x2-6x+8)=3x(x-4)(x-2)
If: y = x2+2x-7 and y = 17-3x Then it follows that: x2+2x-7 = 17-3x Subtract 17 and add 3x to both sides thus forming a quadratic equation: x2+5x-24 = 0 When factored: (x+8)(x-3) = 0 Therefore: x = -8 or x = 3 So when x = -8, y = 41 and when x = 3, y = 8 So the line touches the curve at points A(-8, 41) and B(3, 8) (x1-x2)2 + (y1-y2)2 = length of line segment2 (-8-3)2 + (41-8)2 = 1210 and the square root of this is 34.785 correct to 3dp Length of line segment = 34.785 units
Suppose the middle one is 3x, then the other two are 3x-3 and 3x+3. Their product is (3x-3)*3x*(3x+3) = 3x*(3x-3)*(3x+3) = 27x(x-1)*(x+1) = 27x*(x2-1) So 27x*(x2-1) = 648 x*(x2-1) = 648/27 = 24 ie x3 - x - 24 = 0 or x3 - 3x2 + 3x2 - 9x + 8x - 24 = 0 ie (x - 3)(x2 + 3x + 8) = 0 The only integer solution is x = 3 So the three numbers are 3*(3-1), 3*3 and 3*(3+1) = 6, 9 and 12.
(x + 8)(x - 5)