Solve 6x^2-x-40

$$6 x ^ { 2 } - x - 40$$

Factor

$$\left(3x-8\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-40\right)=-240$$

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 -240.

$$1,-240 2,-120 3,-80 4,-60 5,-48 6,-40 8,-30 10,-24 12,-20 15,-16$$

Calculate the sum for each pair.

$$1-240=-239 2-120=-118 3-80=-77 4-60=-56 5-48=-43 6-40=-34 8-30=-22 10-24=-14 12-20=-8 15-16=-1$$

The solution is the pair that gives sum -1.

$$a=-16 b=15$$

Rewrite 6x^{2}-x-40 as \left(6x^{2}-16x\right)+\left(15x-40\right).

$$\left(6x^{2}-16x\right)+\left(15x-40\right)$$

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

$$2x\left(3x-8\right)+5\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(2x+5\right)$$

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.

$$6x^{2}-x-40=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-40\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-40\right)}}{2\times 6}$$

Multiply -24 times -40.

$$x=\frac{-\left(-1\right)±\sqrt{1+960}}{2\times 6}$$

Add 1 to 960.

$$x=\frac{-\left(-1\right)±\sqrt{961}}{2\times 6}$$

Take the square root of 961.

$$x=\frac{-\left(-1\right)±31}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±31}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±31}{12}$$

Now solve the equation x=\frac{1±31}{12} when ± is plus. Add 1 to 31.

$$x=\frac{32}{12}$$

Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{1±31}{12} when ± is minus. Subtract 31 from 1.

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

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 \frac{8}{3} for x_{1} and -\frac{5}{2} for x_{2}.

$$6x^{2}-x-40=6\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$6x^{2}-x-40=6\left(x-\frac{8}{3}\right)\left(x+\frac{5}{2}\right)$$

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-x-40=6\times \frac{3x-8}{3}\left(x+\frac{5}{2}\right)$$

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

$$6x^{2}-x-40=6\times \frac{3x-8}{3}\times \frac{2x+5}{2}$$

Multiply \frac{3x-8}{3} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-x-40=6\times \frac{\left(3x-8\right)\left(2x+5\right)}{3\times 2}$$

Multiply 3 times 2.

$$6x^{2}-x-40=6\times \frac{\left(3x-8\right)\left(2x+5\right)}{6}$$

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

$$6x^{2}-x-40=\left(3x-8\right)\left(2x+5\right)$$

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

$$x ^ 2 -\frac{1}{6}x -\frac{20}{3} = 0$$

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

$$r + s = \frac{1}{6} rs = -\frac{20}{3}$$

Two numbers r and s sum up to \frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{6} = \frac{1}{12}. 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>

$$r = \frac{1}{12} - u s = \frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{3}

$$(\frac{1}{12} - u) (\frac{1}{12} + u) = -\frac{20}{3}$$

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

$$\frac{1}{144} - u^2 = -\frac{20}{3}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{20}{3}-\frac{1}{144} = -\frac{961}{144}$$

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

$$u^2 = \frac{961}{144} u = \pm\sqrt{\frac{961}{144}} = \pm \frac{31}{12}$$

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

$$r =\frac{1}{12} - \frac{31}{12} = -2.500 s = \frac{1}{12} + \frac{31}{12} = 2.667$$

Evaluate

$$\left(3x-8\right)\left(2x+5\right)$$