Solve -x^2+4x-3

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

Factor

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

Solution Steps

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

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

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

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

Factor out -x in -x^{2}+3x.

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

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

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

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.

$$-x^{2}+4x-3=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-4±\sqrt{16-12}}{2\left(-1\right)}$$

Add 16 to -12.

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

Take the square root of 4.

$$x=\frac{-4±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2}{-2}$$

Now solve the equation x=\frac{-4±2}{-2} when ± is plus. Add -4 to 2.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-4±2}{-2} when ± is minus. Subtract 2 from -4.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

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 3 for x_{2}.

$$-x^{2}+4x-3=-\left(x-1\right)\left(x-3\right)$$

Evaluate

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