Solve F(x)=2x^2+x-15

$$F ( x ) = 2 x ^ { 2 } + x - 15$$

Factor

$$\left(2x-5\right)\left(x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-15\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite 2x^{2}+x-15 as \left(2x^{2}-5x\right)+\left(6x-15\right).

$$\left(2x^{2}-5x\right)+\left(6x-15\right)$$

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

$$x\left(2x-5\right)+3\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+3\right)$$

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.

$$2x^{2}+x-15=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-1±\sqrt{1+120}}{2\times 2}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±11}{4}$$

Now solve the equation x=\frac{-1±11}{4} when ± is plus. Add -1 to 11.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-1±11}{4} when ± is minus. Subtract 11 from -1.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -3 for x_{2}.

$$2x^{2}+x-15=2\left(x-\frac{5}{2}\right)\left(x-\left(-3\right)\right)$$

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

$$2x^{2}+x-15=2\left(x-\frac{5}{2}\right)\left(x+3\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}+x-15=2\times \frac{2x-5}{2}\left(x+3\right)$$

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

$$2x^{2}+x-15=\left(2x-5\right)\left(x+3\right)$$

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

$$x ^ 2 +\frac{1}{2}x -\frac{15}{2} = 0$$

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

$$r + s = -\frac{1}{2} rs = -\frac{15}{2}$$

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{15}{2}

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{15}{2}$$

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

$$\frac{1}{16} - u^2 = -\frac{15}{2}$$

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

$$-u^2 = -\frac{15}{2}-\frac{1}{16} = -\frac{121}{16}$$

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.

$$r =-\frac{1}{4} - \frac{11}{4} = -3 s = -\frac{1}{4} + \frac{11}{4} = 2.500$$

Evaluate

$$\left(2x-5\right)\left(x+3\right)$$