Solve 6x^2-22x-40

$$6 x ^ { 2 } - 22 x - 40$$

Factor

$$2\left(x-5\right)\left(3x+4\right)$$

Solution Steps

Factor out 2.

$$2\left(3x^{2}-11x-20\right)$$

Consider 3x^{2}-11x-20. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\left(-20\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$a=-15 b=4$$

Rewrite 3x^{2}-11x-20 as \left(3x^{2}-15x\right)+\left(4x-20\right).

$$\left(3x^{2}-15x\right)+\left(4x-20\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-5\right)+4\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x+4\right)$$

Rewrite the complete factored expression.

$$2\left(x-5\right)\left(3x+4\right)$$

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.

$$6x^{2}-22x-40=0$$

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.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 6\left(-40\right)}}{2\times 6}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 6\left(-40\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-22\right)±\sqrt{484-24\left(-40\right)}}{2\times 6}$$

Multiply -24 times -40.

$$x=\frac{-\left(-22\right)±\sqrt{484+960}}{2\times 6}$$

Add 484 to 960.

$$x=\frac{-\left(-22\right)±\sqrt{1444}}{2\times 6}$$

Take the square root of 1444.

$$x=\frac{-\left(-22\right)±38}{2\times 6}$$

The opposite of -22 is 22.

$$x=\frac{22±38}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{22±38}{12}$$

Now solve the equation x=\frac{22±38}{12} when ± is plus. Add 22 to 38.

$$x=\frac{60}{12}$$

Divide 60 by 12.

$$x=5$$

Now solve the equation x=\frac{22±38}{12} when ± is minus. Subtract 38 from 22.

$$x=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{4}{3}$$

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 -\frac{4}{3} for x_{2}.

$$6x^{2}-22x-40=6\left(x-5\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$6x^{2}-22x-40=6\left(x-5\right)\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-22x-40=6\left(x-5\right)\times \frac{3x+4}{3}$$

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

$$6x^{2}-22x-40=2\left(x-5\right)\left(3x+4\right)$$

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

$$x ^ 2 -\frac{11}{3}x -\frac{20}{3} = 0$$

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

$$r + s = \frac{11}{3} rs = -\frac{20}{3}$$

Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{6}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{11}{6} - u s = \frac{11}{6} + u$$

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

$$(\frac{11}{6} - u) (\frac{11}{6} + u) = -\frac{20}{3}$$

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

$$\frac{121}{36} - u^2 = -\frac{20}{3}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = -\frac{20}{3}-\frac{121}{36} = -\frac{361}{36}$$

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

$$u^2 = \frac{361}{36} u = \pm\sqrt{\frac{361}{36}} = \pm \frac{19}{6}$$

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

$$r =\frac{11}{6} - \frac{19}{6} = -1.333 s = \frac{11}{6} + \frac{19}{6} = 5$$

Evaluate

$$2\left(x-5\right)\left(3x+4\right)$$