Solve 10-11x-6x^2

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

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

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

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

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

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

Factor out common term -3x+2 by using distributive property.

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

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.

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

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.

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

Square -11.

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

Multiply -4 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+24\times 10}}{2\left(-6\right)}$$

Multiply 24 times 10.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\left(-6\right)}$$

Add 121 to 240.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\left(-6\right)}$$

Take the square root of 361.

$$x=\frac{-\left(-11\right)±19}{2\left(-6\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±19}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{11±19}{-12}$$

Now solve the equation x=\frac{11±19}{-12} when ± is plus. Add 11 to 19.

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

Reduce the fraction \frac{30}{-12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{11±19}{-12} when ± is minus. Subtract 19 from 11.

$$x=-\frac{8}{-12}$$

Reduce the fraction \frac{-8}{-12} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

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

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

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

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

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

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

Multiply -2 times -3.

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

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

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

Evaluate

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