Solve -x^2-3x+54

$$- x ^ { 2 } - 3 x + 54$$

Factor

$$\left(6-x\right)\left(x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+54. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-54=-54$$

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 -54.

$$1,-54 2,-27 3,-18 6,-9$$

Calculate the sum for each pair.

$$1-54=-53 2-27=-25 3-18=-15 6-9=-3$$

The solution is the pair that gives sum -3.

$$a=6 b=-9$$

Rewrite -x^{2}-3x+54 as \left(-x^{2}+6x\right)+\left(-9x+54\right).

$$\left(-x^{2}+6x\right)+\left(-9x+54\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(-x+6\right)+9\left(-x+6\right)$$

Factor out common term -x+6 by using distributive property.

$$\left(-x+6\right)\left(x+9\right)$$

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.

$$-x^{2}-3x+54=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 54}}{2\left(-1\right)}$$

Multiply 4 times 54.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\left(-1\right)}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\left(-1\right)}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±15}{-2}$$

Now solve the equation x=\frac{3±15}{-2} when ± is plus. Add 3 to 15.

$$x=\frac{18}{-2}$$

Divide 18 by -2.

$$x=-9$$

Now solve the equation x=\frac{3±15}{-2} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -9 for x_{1} and 6 for x_{2}.

$$-x^{2}-3x+54=-\left(x-\left(-9\right)\right)\left(x-6\right)$$

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

$$-x^{2}-3x+54=-\left(x+9\right)\left(x-6\right)$$

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.

$$x ^ 2 +3x -54 = 0$$

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

$$r + s = -3 rs = -54$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -54

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -54$$

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

$$\frac{9}{4} - u^2 = -54$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -54-\frac{9}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =-\frac{3}{2} - \frac{15}{2} = -9 s = -\frac{3}{2} + \frac{15}{2} = 6$$

Evaluate

$$\left(6-x\right)\left(x+9\right)$$