Solve -10x^2-11x+6

$$- 10 x ^ { 2 } - 11 x + 6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -10x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=-10\times 6=-60$$

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.

$$a=4 b=-15$$

Rewrite -10x^{2}-11x+6 as \left(-10x^{2}+4x\right)+\left(-15x+6\right).

$$\left(-10x^{2}+4x\right)+\left(-15x+6\right)$$

Factor out 2x in the first and 3 in the second group.

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

Factor out common term -5x+2 by using distributive property.

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

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.

$$-10x^{2}-11x+6=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-10\right)\times 6}}{2\left(-10\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-10\right)\times 6}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-11\right)±\sqrt{121+40\times 6}}{2\left(-10\right)}$$

Multiply 40 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\left(-10\right)}$$

Add 121 to 240.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\left(-10\right)}$$

Take the square root of 361.

$$x=\frac{-\left(-11\right)±19}{2\left(-10\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±19}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{11±19}{-20}$$

Now solve the equation x=\frac{11±19}{-20} when ± is plus. Add 11 to 19.

$$x=\frac{30}{-20}$$

Reduce the fraction \frac{30}{-20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{11±19}{-20} when ± is minus. Subtract 19 from 11.

$$x=-\frac{8}{-20}$$

Reduce the fraction \frac{-8}{-20} to lowest terms by extracting and canceling out 4.

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

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

$$-10x^{2}-11x+6=-10\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\frac{2}{5}\right)$$

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

$$-10x^{2}-11x+6=-10\left(x+\frac{3}{2}\right)\left(x-\frac{2}{5}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-10x^{2}-11x+6=-10\times \frac{-2x-3}{-2}\left(x-\frac{2}{5}\right)$$

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

$$-10x^{2}-11x+6=-10\times \frac{-2x-3}{-2}\times \frac{-5x+2}{-5}$$

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

$$-10x^{2}-11x+6=-10\times \frac{\left(-2x-3\right)\left(-5x+2\right)}{-2\left(-5\right)}$$

Multiply -2 times -5.

$$-10x^{2}-11x+6=-10\times \frac{\left(-2x-3\right)\left(-5x+2\right)}{10}$$

Cancel out 10, the greatest common factor in -10 and 10.

$$-10x^{2}-11x+6=-\left(-2x-3\right)\left(-5x+2\right)$$

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.

$$x ^ 2 +\frac{11}{10}x -\frac{3}{5} = 0$$

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

$$r + s = -\frac{11}{10} rs = -\frac{3}{5}$$

Two numbers r and s sum up to -\frac{11}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{10} = -\frac{11}{20}. 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>

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

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

$$(-\frac{11}{20} - u) (-\frac{11}{20} + u) = -\frac{3}{5}$$

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

$$\frac{121}{400} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{121}{400} on both sides

$$-u^2 = -\frac{3}{5}-\frac{121}{400} = -\frac{361}{400}$$

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

$$u^2 = \frac{361}{400} u = \pm\sqrt{\frac{361}{400}} = \pm \frac{19}{20}$$

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

$$r =-\frac{11}{20} - \frac{19}{20} = -1.500 s = -\frac{11}{20} + \frac{19}{20} = 0.400$$

Evaluate

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