Solve -3x^2-5x+2

$$- 3 x ^ { 2 } - 5 x + 2$$

Factor

$$-\left(3x-1\right)\left(x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-3\times 2=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -3x^{2}-5x+2 as \left(-3x^{2}+x\right)+\left(-6x+2\right).

$$\left(-3x^{2}+x\right)+\left(-6x+2\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(3x-1\right)-2\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-x-2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}-5x+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\left(-3\right)}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2\left(-3\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{5±7}{-6}$$

Now solve the equation x=\frac{5±7}{-6} when ± is plus. Add 5 to 7.

$$x=\frac{12}{-6}$$

Divide 12 by -6.

$$x=-2$$

Now solve the equation x=\frac{5±7}{-6} when ± is minus. Subtract 7 from 5.

$$x=-\frac{2}{-6}$$

Reduce the fraction \frac{-2}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and \frac{1}{3} for x_{2}.

$$-3x^{2}-5x+2=-3\left(x-\left(-2\right)\right)\left(x-\frac{1}{3}\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}-5x+2=-3\left(x+2\right)\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}-5x+2=-3\left(x+2\right)\times \frac{-3x+1}{-3}$$

Cancel out 3, the greatest common factor in -3 and 3.

$$-3x^{2}-5x+2=\left(x+2\right)\left(-3x+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +\frac{5}{3}x -\frac{2}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{5}{3} rs = -\frac{2}{3}$$

Two numbers r and s sum up to -\frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{3} = -\frac{5}{6}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{5}{6} - u s = -\frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}

$$(-\frac{5}{6} - u) (-\frac{5}{6} + u) = -\frac{2}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{36} - u^2 = -\frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = -\frac{2}{3}-\frac{25}{36} = -\frac{49}{36}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{5}{6} - \frac{7}{6} = -2 s = -\frac{5}{6} + \frac{7}{6} = 0.333$$

Evaluate

$$-\left(3x-1\right)\left(x+2\right)$$