Solve 8+2x-3x^2=

$$8 + 2 x - 3 x ^ { 2 } =$$

Factor

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

Solution Steps

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

$$-3x^{2}+2x+8$$

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3\times 8=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 2.

$$a=6 b=-4$$

Rewrite -3x^{2}+2x+8 as \left(-3x^{2}+6x\right)+\left(-4x+8\right).

$$\left(-3x^{2}+6x\right)+\left(-4x+8\right)$$

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

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

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

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

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.

$$-3x^{2}+2x+8=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12\times 8}}{2\left(-3\right)}$$

Multiply 12 times 8.

$$x=\frac{-2±\sqrt{4+96}}{2\left(-3\right)}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\left(-3\right)}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±10}{-6}$$

Now solve the equation x=\frac{-2±10}{-6} when ± is plus. Add -2 to 10.

$$x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-2±10}{-6} when ± is minus. Subtract 10 from -2.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

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

$$-3x^{2}+2x+8=-3\left(x-\left(-\frac{4}{3}\right)\right)\left(x-2\right)$$

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

$$-3x^{2}+2x+8=-3\left(x+\frac{4}{3}\right)\left(x-2\right)$$

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

$$-3x^{2}+2x+8=-3\times \frac{-3x-4}{-3}\left(x-2\right)$$

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

$$-3x^{2}+2x+8=\left(-3x-4\right)\left(x-2\right)$$

Evaluate

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