Solve -6x^2-x+15

$$- 6 x ^ { 2 } - x + 15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6\times 15=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -1.

$$a=9 b=-10$$

Rewrite -6x^{2}-x+15 as \left(-6x^{2}+9x\right)+\left(-10x+15\right).

$$\left(-6x^{2}+9x\right)+\left(-10x+15\right)$$

Factor out -3x in the first and -5 in the second group.

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

Factor out common term 2x-3 by using distributive property.

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

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.

$$-6x^{2}-x+15=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)\times 15}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24\times 15}}{2\left(-6\right)}$$

Multiply 24 times 15.

$$x=\frac{-\left(-1\right)±\sqrt{1+360}}{2\left(-6\right)}$$

Add 1 to 360.

$$x=\frac{-\left(-1\right)±\sqrt{361}}{2\left(-6\right)}$$

Take the square root of 361.

$$x=\frac{-\left(-1\right)±19}{2\left(-6\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±19}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{1±19}{-12}$$

Now solve the equation x=\frac{1±19}{-12} when ± is plus. Add 1 to 19.

$$x=\frac{20}{-12}$$

Reduce the fraction \frac{20}{-12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{3}$$

Now solve the equation x=\frac{1±19}{-12} when ± is minus. Subtract 19 from 1.

$$x=-\frac{18}{-12}$$

Reduce the fraction \frac{-18}{-12} to lowest terms by extracting and canceling out 6.

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

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

$$-6x^{2}-x+15=-6\left(x-\left(-\frac{5}{3}\right)\right)\left(x-\frac{3}{2}\right)$$

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

$$-6x^{2}-x+15=-6\left(x+\frac{5}{3}\right)\left(x-\frac{3}{2}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}-x+15=-6\times \frac{-3x-5}{-3}\left(x-\frac{3}{2}\right)$$

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

$$-6x^{2}-x+15=-6\times \frac{-3x-5}{-3}\times \frac{-2x+3}{-2}$$

Multiply \frac{-3x-5}{-3} times \frac{-2x+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}-x+15=-6\times \frac{\left(-3x-5\right)\left(-2x+3\right)}{-3\left(-2\right)}$$

Multiply -3 times -2.

$$-6x^{2}-x+15=-6\times \frac{\left(-3x-5\right)\left(-2x+3\right)}{6}$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6x^{2}-x+15=-\left(-3x-5\right)\left(-2x+3\right)$$

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.

$$x ^ 2 +\frac{1}{6}x -\frac{5}{2} = 0$$

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

$$r + s = -\frac{1}{6} rs = -\frac{5}{2}$$

Two numbers r and s sum up to -\frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{6} = -\frac{1}{12}. 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>

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -\frac{5}{2}$$

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

$$\frac{1}{144} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{5}{2}-\frac{1}{144} = -\frac{361}{144}$$

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

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

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

$$r =-\frac{1}{12} - \frac{19}{12} = -1.667 s = -\frac{1}{12} + \frac{19}{12} = 1.500$$

Evaluate

$$15-x-6x^{2}$$