Solve 12-5x-2x^2

$$12 - 5 x - 2 x ^ { 2 }$$

Factor

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

Solution Steps

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

$$-2x^{2}-5x+12$$

Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-2\times 12=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=3 b=-8$$

Rewrite -2x^{2}-5x+12 as \left(-2x^{2}+3x\right)+\left(-8x+12\right).

$$\left(-2x^{2}+3x\right)+\left(-8x+12\right)$$

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

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

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

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

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.

$$-2x^{2}-5x+12=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 12}}{2\left(-2\right)}$$

Multiply 8 times 12.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\left(-2\right)}$$

Add 25 to 96.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of 121.

$$x=\frac{-\left(-5\right)±11}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±11}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±11}{-4}$$

Now solve the equation x=\frac{5±11}{-4} when ± is plus. Add 5 to 11.

$$x=\frac{16}{-4}$$

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{5±11}{-4} when ± is minus. Subtract 11 from 5.

$$x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and \frac{3}{2} for x_{2}.

$$-2x^{2}-5x+12=-2\left(x-\left(-4\right)\right)\left(x-\frac{3}{2}\right)$$

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

$$-2x^{2}-5x+12=-2\left(x+4\right)\left(x-\frac{3}{2}\right)$$

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

$$-2x^{2}-5x+12=-2\left(x+4\right)\times \frac{-2x+3}{-2}$$

Cancel out 2, the greatest common factor in -2 and 2.

$$-2x^{2}-5x+12=\left(x+4\right)\left(-2x+3\right)$$

Evaluate

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