Solve 3x^2-6x-45

$$3 x ^ { 2 } - 6 x - 45$$

Factor

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

Solution Steps

Factor out 3.

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

Consider x^{2}-2x-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.

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

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 -15.

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

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$3x^{2}-6x-45=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-45\right)}}{2\times 3}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-45\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\left(-45\right)}}{2\times 3}$$

Multiply -12 times -45.

$$x=\frac{-\left(-6\right)±\sqrt{36+540}}{2\times 3}$$

Add 36 to 540.

$$x=\frac{-\left(-6\right)±\sqrt{576}}{2\times 3}$$

Take the square root of 576.

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

The opposite of -6 is 6.

$$x=\frac{6±24}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±24}{6}$$

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

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

Divide 30 by 6.

$$x=5$$

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

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

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}.

$$3x^{2}-6x-45=3\left(x-5\right)\left(x-\left(-3\right)\right)$$

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

$$3x^{2}-6x-45=3\left(x-5\right)\left(x+3\right)$$

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.This is achieved by dividing both sides of the equation by 3

$$x ^ 2 -2x -15 = 0$$

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

$$r + s = 2 rs = -15$$

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

$$r = 1 - u s = 1 + u$$

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

$$(1 - u) (1 + u) = -15$$

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

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

Simplify the expression by subtracting 1 on both sides

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

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

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

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

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

Evaluate

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