Solve (p^2-4p+3)

$$( p ^ { 2 } - 4 p + 3 )$$

Factor

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

Solution Steps

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

$$a+b=-4 ab=1\times 3=3$$

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

$$a=-3 b=-1$$

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

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

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

$$p\left(p-3\right)-\left(p-3\right)$$

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

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

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.

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

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.

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

Square -4.

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

Multiply -4 times 3.

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

Add 16 to -12.

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

Take the square root of 4.

$$p=\frac{-\left(-4\right)±2}{2}$$

The opposite of -4 is 4.

$$p=\frac{4±2}{2}$$

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

$$p=\frac{6}{2}$$

Divide 6 by 2.

$$p=3$$

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

$$p=\frac{2}{2}$$

Divide 2 by 2.

$$p=1$$

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

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

Evaluate

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