Solve 6x^2+4x-10

$$6 x ^ { 2 } + 4 x - 10$$

Factor

$$2\left(x-1\right)\left(3x+5\right)$$

Solution Steps

Factor out 2.

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

Consider 3x^{2}+2x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-5\right)=-15$$

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 -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.

$$a=-3 b=5$$

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

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

Factor out 3x in the first and 5 in the second group.

$$3x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x+5\right)$$

Rewrite the complete factored expression.

$$2\left(x-1\right)\left(3x+5\right)$$

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.

$$6x^{2}+4x-10=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6\left(-10\right)}}{2\times 6}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 6\left(-10\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-4±\sqrt{16-24\left(-10\right)}}{2\times 6}$$

Multiply -24 times -10.

$$x=\frac{-4±\sqrt{16+240}}{2\times 6}$$

Add 16 to 240.

$$x=\frac{-4±\sqrt{256}}{2\times 6}$$

Take the square root of 256.

$$x=\frac{-4±16}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-4±16}{12}$$

Now solve the equation x=\frac{-4±16}{12} when ± is plus. Add -4 to 16.

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

Divide 12 by 12.

$$x=1$$

Now solve the equation x=\frac{-4±16}{12} when ± is minus. Subtract 16 from -4.

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

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{3}$$

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

$$6x^{2}+4x-10=6\left(x-1\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$6x^{2}+4x-10=6\left(x-1\right)\left(x+\frac{5}{3}\right)$$

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

$$6x^{2}+4x-10=6\left(x-1\right)\times \frac{3x+5}{3}$$

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

$$6x^{2}+4x-10=2\left(x-1\right)\left(3x+5\right)$$

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

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

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

$$r + s = -\frac{2}{3} rs = -\frac{5}{3}$$

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

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -\frac{5}{3}$$

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

$$\frac{1}{9} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{5}{3}-\frac{1}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-\frac{1}{3} - \frac{4}{3} = -1.667 s = -\frac{1}{3} + \frac{4}{3} = 1.000$$

Evaluate

$$2\left(x-1\right)\left(3x+5\right)$$