Solve 4-x^2+3x+2x

$$4 - x ^ { 2 } + 3 x + 2 x$$

Factor

$$-\left(x-\frac{5-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+5}{2}\right)$$

Steps Using the Quadratic Formula

Combine 3x and 2x to get 5x.

$$factor(4-x^{2}+5x)$$

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.

$$-x^{2}+5x+4=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-5±\sqrt{25+16}}{2\left(-1\right)}$$

Add 25 to 16.

$$x=\frac{-5±\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{41}}{-2}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{-2} when ± is plus. Add -5 to \sqrt{41}.

$$x=\frac{\sqrt{41}-5}{-2}$$

Divide -5+\sqrt{41} by -2.

$$x=\frac{5-\sqrt{41}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{-2} when ± is minus. Subtract \sqrt{41} from -5.

$$x=\frac{-\sqrt{41}-5}{-2}$$

Divide -5-\sqrt{41} by -2.

$$x=\frac{\sqrt{41}+5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5-\sqrt{41}}{2} for x_{1} and \frac{5+\sqrt{41}}{2} for x_{2}.

$$-x^{2}+5x+4=-\left(x-\frac{5-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+5}{2}\right)$$

Evaluate

$$4+5x-x^{2}$$

Solution Steps

Combine 3x and 2x to get 5x.

$$4-x^{2}+5x$$