# SOLUTION: solve for x (3x-2)(x+5)=4x+2

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 Click here to see ALL problems on Linear Algebra Question 35903: solve for x (3x-2)(x+5)=4x+2Answer by atif.muhammad(135)   (Show Source): You can put this solution on YOUR website!```x (3x-2)(x+5)=4x+2 Let's open up one of the brackets first x(3x^2 + 15x - 2x - 10) = 4x + 2 Let's open up the last bracket 3x^3 + 15x^2 - 2x^2 - 10x = 4x + 2 Now let's move all the variables to the left hand side 3x^3 + 13x^2 -14x - 2 = 0 Now we have a cubic equation. To solve this, we use the factor theorem. f(x) = 3x^3 + 13x^2 -14x - 2 When f(a) = 0, then (x-a) is a factor of f(x) f(0) = -2 f(-1) = -3 + 13 + 14 -2 = 22 f(1) = 3 + 13 -14 -2 = 0 f(1) = 0, therefore (x-1) is a factor of f(x) We now need to divide f(x) by (x-1) and find a quadratic equation. I have 'magically' divided it. (using the long division method) f(x) = (3x^2 + 16x + 2)(x-1) We now have a quadratic and a linear. (3x^2 + 16x + 2)(x-1) = 0 Solve linear: x-1 = 0 x = 1 --> First solution Solve quadratic: 3x^2 + 16x + 2 = 0 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=232 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -0.128075631378697, -5.20525770195464. Here's your graph: We now know our 3 solutions! x= 1, -5.21, -0.128```