Solve -3x^2-24x-45

$$- 3 x ^ { 2 } - 24 x - 45$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(-x^{2}-8x-15\right)$$

Consider -x^{2}-8x-15. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-\left(-15\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-3 b=-5$$

Rewrite -x^{2}-8x-15 as \left(-x^{2}-3x\right)+\left(-5x-15\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$-3x^{2}-24x-45=0$$

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.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-3\right)\left(-45\right)}}{2\left(-3\right)}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\left(-3\right)\left(-45\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-24\right)±\sqrt{576+12\left(-45\right)}}{2\left(-3\right)}$$

Multiply 12 times -45.

$$x=\frac{-\left(-24\right)±\sqrt{576-540}}{2\left(-3\right)}$$

Add 576 to -540.

$$x=\frac{-\left(-24\right)±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{-\left(-24\right)±6}{2\left(-3\right)}$$

The opposite of -24 is 24.

$$x=\frac{24±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{24±6}{-6}$$

Now solve the equation x=\frac{24±6}{-6} when ± is plus. Add 24 to 6.

$$x=\frac{30}{-6}$$

Divide 30 by -6.

$$x=-5$$

Now solve the equation x=\frac{24±6}{-6} when ± is minus. Subtract 6 from 24.

$$x=\frac{18}{-6}$$

Divide 18 by -6.

$$x=-3$$

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

$$-3x^{2}-24x-45=-3\left(x-\left(-5\right)\right)\left(x-\left(-3\right)\right)$$

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

$$-3x^{2}-24x-45=-3\left(x+5\right)\left(x+3\right)$$

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.

$$x ^ 2 +8x +15 = 0$$

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

$$r + s = -8 rs = 15$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(-4 - u) (-4 + u) = 15$$

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

$$16 - u^2 = 15$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 15-16 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =-4 - 1 = -5 s = -4 + 1 = -3$$

Evaluate

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