Solve g(x)=2x^2-5x-7

$$g ( x ) = 2 x ^ { 2 } - 5 x - 7$$

Factor

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

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.

$$a+b=-5 ab=2\left(-7\right)=-14$$

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.

$$a=-7 b=2$$

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

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

Factor out x in 2x^{2}-7x.

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

Factor out common term 2x-7 by using distributive property.

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

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.

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

Square -5.

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

Multiply -4 times 2.

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

Multiply -8 times -7.

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

Add 25 to 56.

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

Take the square root of 81.

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

The opposite of -5 is 5.

$$x=\frac{5±9}{2\times 2}$$

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}.

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

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

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

Subtract \frac{7}{2} from x by finding a common denominator and subtracting 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.This is achieved by dividing both sides of the equation by 2

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

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.

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

Evaluate

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