factor n^2-4n+3 — step-by-step solution

Let's factor in

ℝ

n^{2} - 4 n + 3

Given the quadratic form

a n^{2} + b n + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -4

and

c = 3

Δ = {(-4)}^{2} - 4 \cdot 1 \cdot 3

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-4)}^{2}

(-4) \cdot (-4)

16

Δ = 16 - 4 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 16 - 12

ⓘ Start from

16

and move back

12

on the number line:

16 - 12

4

Δ = 4

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

n_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-4) - \sqrt{4}}{2 \cdot 1}

n_{1} = \frac{4 - \sqrt{4}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{4}

2

;

2^{2}

4

n_{1} = \frac{4 - 2}{2}

ⓘ The numerator and the denominator are computed before the division:

4 - 2

2

n_{1} = \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

n_{1} = 1

The quadratic formula gives

n_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-4) + \sqrt{4}}{2 \cdot 1}

n_{2} = \frac{4 + \sqrt{4}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{4}

2

;

2^{2}

4

n_{2} = \frac{4 + 2}{2}

ⓘ The numerator and the denominator are computed before the division:

4 + 2

6

n_{2} = \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

n_{2} = 3

▶Factors identified:

(n - 1) \cdot (n - 3)