Solve 2m^2-3m-14

$$2 m ^ { 2 } - 3 m - 14$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2m^{2}+am+bm-14. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-14\right)=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite 2m^{2}-3m-14 as \left(2m^{2}-7m\right)+\left(4m-14\right).

$$\left(2m^{2}-7m\right)+\left(4m-14\right)$$

Factor out m in the first and 2 in the second group.

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

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

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

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.

$$2m^{2}-3m-14=0$$

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.

$$m=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-14\right)}}{2\times 2}$$

Square -3.

$$m=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-14\right)}}{2\times 2}$$

Multiply -4 times 2.

$$m=\frac{-\left(-3\right)±\sqrt{9-8\left(-14\right)}}{2\times 2}$$

Multiply -8 times -14.

$$m=\frac{-\left(-3\right)±\sqrt{9+112}}{2\times 2}$$

Add 9 to 112.

$$m=\frac{-\left(-3\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$m=\frac{-\left(-3\right)±11}{2\times 2}$$

The opposite of -3 is 3.

$$m=\frac{3±11}{2\times 2}$$

Multiply 2 times 2.

$$m=\frac{3±11}{4}$$

Now solve the equation m=\frac{3±11}{4} when ± is plus. Add 3 to 11.

$$m=\frac{14}{4}$$

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

$$m=\frac{7}{2}$$

Now solve the equation m=\frac{3±11}{4} when ± is minus. Subtract 11 from 3.

$$m=-\frac{8}{4}$$

Divide -8 by 4.

$$m=-2$$

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 -2 for x_{2}.

$$2m^{2}-3m-14=2\left(m-\frac{7}{2}\right)\left(m-\left(-2\right)\right)$$

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

$$2m^{2}-3m-14=2\left(m-\frac{7}{2}\right)\left(m+2\right)$$

Subtract \frac{7}{2} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2m^{2}-3m-14=2\times \frac{2m-7}{2}\left(m+2\right)$$

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

$$2m^{2}-3m-14=\left(2m-7\right)\left(m+2\right)$$

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

$$x ^ 2 -\frac{3}{2}x -7 = 0$$

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

$$r + s = \frac{3}{2} rs = -7$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -7

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -7$$

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

$$\frac{9}{16} - u^2 = -7$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -7-\frac{9}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =\frac{3}{4} - \frac{11}{4} = -2 s = \frac{3}{4} + \frac{11}{4} = 3.500$$

Evaluate

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