3(x + 1)(x + 2)
x2+9x+18 find factors of 18 that add up to 9, in this case 6&3 (x+6)(x+3)
3x2 + 15x + 6 = 3*(x2 + 5x + 2). There are no further rational factors.
Asuming you ment The Factor of x^2+9x+18 the answer is (x+3)(x+6) Asuming you ment The Factor of x^2+9x+18 the answer is (x+3)(x+6)
2x + 6 + 9x = 39 11x + 6 = 39 11x = 39 - 6 11x = 33 x = 3
9x-4 = 7x+8 9x-7x = 8+4 2x = 12 x = 6
3x2+7-6 = (3x-2)(x+3) when factored
(3x + 1)(x + 6)
x^2-9x+18=(x-6)(x-3)
(x2 - 2)(y - 3)
x2+9x+18 find factors of 18 that add up to 9, in this case 6&3 (x+6)(x+3)
x2 - 9x + 18 = (x - 3)(x - 6)
3x2 + 15x + 6 = 3*(x2 + 5x + 2). There are no further rational factors.
(x - 6)(x - 3)
9x+6 = 7x+18 9x-7x = 18-6 2x = 12 x = 6
3x2=6+38=44
3x2 - 9x = -5; add 5 to both sides, in order to use the quadratic formula 3x2 - 9x + 5 = 0; a = 3, b = -9, and c = 5 x = [-b ± √(b2 - 4ac)]/(2a) x = {-(-9) ± √[(-9)2 - 4(3)(5)]}/[(2)(3)] = [9 ± √(81 - 60)]/6 = [9 ± √21)/6
3x^(2) +9x - 2x -6 Collect 'like terms'. Hence 3x^(2) + 7x - 6 Next write down all the factors of '3' and '6' Hence 3 ; 1' & 3' 6 ; 1,6 ; 2,3. From these pairs of number we select a pair from each coefficient, that add/multiply to '7' . Hence (3' x 3 ) & (1' x 2) ; NB 'dashes' (') to indicate source of numbers. Write up brackets (3x 2)(x 3) -2)(x + 3) Next we notice that the '6' is negative, so the two signs are different (+/-) The '7x' is positive , so the larger number takes the positive sign . Hence (3x - 2)(x + 3)