6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
This equation can't actually be factored, but you can solve for x: 6x2 + 37x + 6 = 0 ∴ 6x2 + 37x = -6 ∴ x2 + (37/6)x = -1 ∴ x2 + (37/6)x + (37/12)2 = -1 - (37/12)2 ∴ (x + 37/12)2 = -1 - (37/12)2 ∴ x + 37/12 = ±[-1 - (37/12)2]1/2 ∴ x = 37/12 ±[-1 - (37/12)2]1/2 Which gives you two complex numbers for the answer, approximately: 37 / 12 + 3.2414417 × i 37 / 12 - 3.2414417 × i
(6 × 2) + 3 = 15
6x2 + 10x = 5 6x2 + 10x - 5 = 0 The roots are 1/(2*6)* [-10 +/- sqrt(102 - 4*6*(-5))] = 1/(2*6) * [-10 +/- sqrt 220] =1/6 * [-5 +/- sqrt 55] So x = -2.0694 or 0.403
6(x - 2)(x - 1)
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
This equation can't actually be factored, but you can solve for x: 6x2 + 37x + 6 = 0 ∴ 6x2 + 37x = -6 ∴ x2 + (37/6)x = -1 ∴ x2 + (37/6)x + (37/12)2 = -1 - (37/12)2 ∴ (x + 37/12)2 = -1 - (37/12)2 ∴ x + 37/12 = ±[-1 - (37/12)2]1/2 ∴ x = 37/12 ±[-1 - (37/12)2]1/2 Which gives you two complex numbers for the answer, approximately: 37 / 12 + 3.2414417 × i 37 / 12 - 3.2414417 × i
6x2-18x+12 = (6x-6)(x-2)
(6 × 2) + 3 = 15
6x2 + 14x -12 = (3x - 2) (2x + 6). There they are.
6x2 + 10x = 5 6x2 + 10x - 5 = 0 The roots are 1/(2*6)* [-10 +/- sqrt(102 - 4*6*(-5))] = 1/(2*6) * [-10 +/- sqrt 220] =1/6 * [-5 +/- sqrt 55] So x = -2.0694 or 0.403
34+(6*2)=34+12+46
6(x - 2)(x - 1)
x3 + 6x2 - 4x - 24 = (x + 6)(x2 - 4) = (x + 6)(x + 2)(x - 2)
x5+4x4-6x2+nx+2 when divided by x+2 has a remainder of 6 Using the remainder theorem: n = 2
6x2-9x-6 is equal to 66.
5 + 6x² + 3x - 7x³ + x6 would have the degree of 6.