(2x+4)(x+1)
quadratic
-4x2 + 6x + 40 = 0 ∴ -2x2 + 3x + 20 = 0 ∴ -2x2 - 8x + 5x + 20 = 0 ∴ -2x(x + 4) + 5(x + 4) = 0 ∴ (5 - 2x)(x + 4) = 0 ∴ x ∈ {5/2, -4}
(x + 2)(x^2+4)
2X2+5x-12 4+5x-12 -8+5x 5x-8
X - 4
Trinomial
quadratic
Not factorable
(2x + 1)(x + 4)
6x(3x2 - x + 4)
x2 + 6x + 8 = (x + 4)(x + 2)
-4x2 + 6x + 40 = 0 ∴ -2x2 + 3x + 20 = 0 ∴ -2x2 - 8x + 5x + 20 = 0 ∴ -2x(x + 4) + 5(x + 4) = 0 ∴ (5 - 2x)(x + 4) = 0 ∴ x ∈ {5/2, -4}
2x^(2) - 5x - 12 First write down all the factors of '2' & '12' 2 = 1,2 12 = 1,12; 2,6; 3,4 . From this we select two pairs of numbers, one pair from each coefficient, that add/multiply to '5' They are, 1,2 & 3,4 Hence -(2 x 4) + (1 x 3) = -8 + 3 = -5 Set up the brackets (2x 3)(x 4) Hence the signs are ( 2x + 3)(x - 4)
x2 + 6x + 8 =(x + 2)(x +4)
(x + 2)(x^2+4)
2(3x^2 + 6x + 2)
2X2+5x-12 4+5x-12 -8+5x 5x-8