Solve 6x^2+8x-30

$$6 x ^ { 2 } + 8 x - 30$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(3x^{2}+4x-15\right)$$

Consider 3x^{2}+4x-15. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-15\right)=-45$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

Rewrite 3x^{2}+4x-15 as \left(3x^{2}-5x\right)+\left(9x-15\right).

$$\left(3x^{2}-5x\right)+\left(9x-15\right)$$

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

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

Factor out common term 3x-5 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}+8x-30=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 6\left(-30\right)}}{2\times 6}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 6\left(-30\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-8±\sqrt{64-24\left(-30\right)}}{2\times 6}$$

Multiply -24 times -30.

$$x=\frac{-8±\sqrt{64+720}}{2\times 6}$$

Add 64 to 720.

$$x=\frac{-8±\sqrt{784}}{2\times 6}$$

Take the square root of 784.

$$x=\frac{-8±28}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-8±28}{12}$$

Now solve the equation x=\frac{-8±28}{12} when ± is plus. Add -8 to 28.

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

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{-8±28}{12} when ± is minus. Subtract 28 from -8.

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

Divide -36 by 12.

$$x=-3$$

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 -3 for x_{2}.

$$6x^{2}+8x-30=6\left(x-\frac{5}{3}\right)\left(x-\left(-3\right)\right)$$

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

$$6x^{2}+8x-30=6\left(x-\frac{5}{3}\right)\left(x+3\right)$$

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

$$6x^{2}+8x-30=6\times \frac{3x-5}{3}\left(x+3\right)$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}+8x-30=2\left(3x-5\right)\left(x+3\right)$$

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 6

$$x ^ 2 +\frac{4}{3}x -5 = 0$$

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

$$r + s = -\frac{4}{3} rs = -5$$

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -5$$

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

$$\frac{4}{9} - u^2 = -5$$

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

$$-u^2 = -5-\frac{4}{9} = -\frac{49}{9}$$

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

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

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

$$r =-\frac{2}{3} - \frac{7}{3} = -3 s = -\frac{2}{3} + \frac{7}{3} = 1.667$$

Evaluate

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