x - 3
(x + 4)(x + 9)
x2 + 11x + 18 (x + 9)(x + 2) CHECK: x2 + 9x + 2x + 18 x2 + 11x + 18 SET EACH EQUAL TO ZERO: x + 9 = 0 x = -9 x + 2 = 0 x = -2 NOW YOU ARE DONE: Solution set: {-9, -2}
No, as it has two factors, (x - 6)(x + 7)
x2 + 48x + 320
x2+24x+144 = 9 x2+24x+144-9 = 0 x2+24x+135 = 0 (x+9)(x+15) = 0 x = -9 or x = -15
It is: (x+1)(x+4) none of which are prime numbers
x2 + 9 = 10x x2 - 10x + 9 = 0. (x - 9)(x - 1) = 0. Therefore, x = 1 or 9.
x2 + 12x + 27 = (x + 9)(x + 3)
x2-6x+9 = (x-3)(x-3) when factorised.
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
X2 or 2 times x?
(x+1)(x+9)
x2 + 13x + 36 = (x + 9)(x + 4)
x2+11x+11 = 7x+9 x2+11x-7x+11-9 = 0 x2+4x+2 = 0 The above quadratic equation can be solved by using the quadratic equation formula and it will have two solutions.
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)