Solve -2x^2-5x+7

$$- 2 x ^ { 2 } - 5 x + 7$$

Factor

$$\left(1-x\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-2\times 7=-14$$

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 -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=2 b=-7$$

Rewrite -2x^{2}-5x+7 as \left(-2x^{2}+2x\right)+\left(-7x+7\right).

$$\left(-2x^{2}+2x\right)+\left(-7x+7\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(-x+1\right)+7\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(2x+7\right)$$

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.

$$-2x^{2}-5x+7=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 7}}{2\left(-2\right)}$$

Multiply 8 times 7.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\left(-2\right)}$$

Add 25 to 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\left(-2\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-5\right)±9}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±9}{-4}$$

Now solve the equation x=\frac{5±9}{-4} when ± is plus. Add 5 to 9.

$$x=\frac{14}{-4}$$

Reduce the fraction \frac{14}{-4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{2}$$

Now solve the equation x=\frac{5±9}{-4} when ± is minus. Subtract 9 from 5.

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

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{7}{2} for x_{1} and 1 for x_{2}.

$$-2x^{2}-5x+7=-2\left(x-\left(-\frac{7}{2}\right)\right)\left(x-1\right)$$

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

$$-2x^{2}-5x+7=-2\left(x+\frac{7}{2}\right)\left(x-1\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-2x^{2}-5x+7=-2\times \frac{-2x-7}{-2}\left(x-1\right)$$

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

$$-2x^{2}-5x+7=\left(-2x-7\right)\left(x-1\right)$$

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.

$$x ^ 2 +\frac{5}{2}x -\frac{7}{2} = 0$$

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

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

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = -\frac{7}{2}$$

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

$$\frac{25}{16} - u^2 = -\frac{7}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{7}{2}-\frac{25}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{5}{4} - \frac{9}{4} = -3.500 s = -\frac{5}{4} + \frac{9}{4} = 1$$

Evaluate

$$\left(1-x\right)\left(2x+7\right)$$