Solve -3x^2-x+2

$$-3 { x }^{ 2 } -x+2$$

Factor

$$-\left(3x-2\right)\left(x+1\right)$$

Solution Steps

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

$$a+b=-1 ab=-3\times 2=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

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

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

Factor out -x in the first and -1 in the second group.

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

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

$$\left(3x-2\right)\left(-x-1\right)$$

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.

$$-3x^{2}-x+2=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-3\right)}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-3\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-3\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{1±5}{-6}$$

Now solve the equation x=\frac{1±5}{-6} when ± is plus. Add 1 to 5.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{1±5}{-6} when ± is minus. Subtract 5 from 1.

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

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

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

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

$$-3x^{2}-x+2=-3\left(x-\left(-1\right)\right)\left(x-\frac{2}{3}\right)$$

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

$$-3x^{2}-x+2=-3\left(x+1\right)\left(x-\frac{2}{3}\right)$$

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

$$-3x^{2}-x+2=-3\left(x+1\right)\times \frac{-3x+2}{-3}$$

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

$$-3x^{2}-x+2=\left(x+1\right)\left(-3x+2\right)$$

Evaluate

$$-\left(3x-2\right)\left(x+1\right)$$