Solve f(x)=-2x^2-x+3

$$f ( x ) = - 2 x ^ { 2 } - x + 3$$

Factor

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

Solution Steps

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

$$a+b=-1 ab=-2\times 3=-6$$

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 -1.

$$a=2 b=-3$$

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

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

Factor out 2x in the first and 3 in the second group.

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

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

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

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.

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

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-2\right)}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±5}{-4}$$

Now solve the equation x=\frac{1±5}{-4} when ± is plus. Add 1 to 5.

$$x=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{1±5}{-4} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

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

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

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

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

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in -2 and 2.

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

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.

$$x ^ 2 +\frac{1}{2}x -\frac{3}{2} = 0$$

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

$$r + s = -\frac{1}{2} rs = -\frac{3}{2}$$

Two numbers r and s sum up to -\frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{2} = -\frac{1}{4}. 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>

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

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

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{3}{2}$$

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

$$\frac{1}{16} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{3}{2}-\frac{1}{16} = -\frac{25}{16}$$

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

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

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

$$r =-\frac{1}{4} - \frac{5}{4} = -1.500 s = -\frac{1}{4} + \frac{5}{4} = 1$$

Evaluate

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