x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x+3) -2(x+3) = 0 so (x+3)(x-2) = 0 so x+3 = 0 or x-2 = 0 so that x = -3 or x = 2
5x2 + 15x = 0 5x(x + 3) = 0 therefore, x = 0 or x + 3 = 0 that is, x = 0 or x = -3
x + x2 = 6 So x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x + 3) - 2(x + 3) = 0 (x + 3)(x - 2) = 0 So x + 3 = 0 so that x = -3 or x - 2 = 0 so that x = 2
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
-4
x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x+3) -2(x+3) = 0 so (x+3)(x-2) = 0 so x+3 = 0 or x-2 = 0 so that x = -3 or x = 2
5x2 + 15x = 0 5x(x + 3) = 0 therefore, x = 0 or x + 3 = 0 that is, x = 0 or x = -3
An equation with the solution set 1 and 3 can be written in factored form as (x-1)(x-3) = 0. When expanded, this equation becomes x^2 - 4x + 3 = 0. Therefore, the equation x^2 - 4x + 3 = 0 has the solution set 1 and 3.
If a polynomial has factors x-6 and x-3 it will equal 0 if either factor equals 0 since the other factor then would be multiplied by 0. ie. 0 * (x-6)=0 and 0 * (x-3)=0. so x=3 or 6
2x2-x-3 = 0= 4 - x - 3 = 0= 1 - x = 0= 1 = xTherefore x = 1
I'll assume that's x(x - 3) - 10(x - 3) = 0 You can factor an x - 3 out of both of those. x - 10 = 0 x = 10 Check it. 10(10 - 3) - 10(10 - 3) = 0 70 - 70 = 0 It checks.
x + x2 = 6 So x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x + 3) - 2(x + 3) = 0 (x + 3)(x - 2) = 0 So x + 3 = 0 so that x = -3 or x - 2 = 0 so that x = 2
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
-4
x2-2x-3 = 0 (x+1)(x-3) = 0 x = -1 or x = 3
x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
2x2-x-15 = 0 When factorised: (2x+5)(x-3) = 0