Solve 6x^2-3x-45

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

Factor

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

Solution Steps

Factor out 3.

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

Consider 2x^{2}-x-15. 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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-6 b=5$$

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

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

Factor out 2x in the first and 5 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6\left(-45\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-24\left(-45\right)}}{2\times 6}$$

Multiply -24 times -45.

$$x=\frac{-\left(-3\right)±\sqrt{9+1080}}{2\times 6}$$

Add 9 to 1080.

$$x=\frac{-\left(-3\right)±\sqrt{1089}}{2\times 6}$$

Take the square root of 1089.

$$x=\frac{-\left(-3\right)±33}{2\times 6}$$

The opposite of -3 is 3.

$$x=\frac{3±33}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{3±33}{12}$$

Now solve the equation x=\frac{3±33}{12} when ± is plus. Add 3 to 33.

$$x=\frac{36}{12}$$

Divide 36 by 12.

$$x=3$$

Now solve the equation x=\frac{3±33}{12} when ± is minus. Subtract 33 from 3.

$$x=-\frac{30}{12}$$

Reduce the fraction \frac{-30}{12} to lowest terms by extracting and canceling out 6.

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

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

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

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

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

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

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

Cancel out 2, the greatest common factor in 6 and 2.

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

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 6

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

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.

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

Evaluate

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