factor 15x^2-13x+2 — step-by-step solution

Let's factor in

ℝ

15 x^{2} - 13 x + 2

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 15

b = -13

and

c = 2

Δ = {(-13)}^{2} - 4 \cdot 15 \cdot 2

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

{(-13)}^{2}

(-13) \cdot (-13)

169

■Multiplications are done before additions and subtractions:

4 \cdot 15

4 \cdot (10 + 5)

4 \cdot 10 + 4 \cdot 5

40 + 20

60

Δ = 169 - 60 \cdot 2

ⓘ Multiplications are done before additions and subtractions:

60 \cdot 2

120

Δ = 169 - 120

ⓘ Start from

169

and move back

120

on the number line:

169 - 120

49

Δ = 49

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-13) - \sqrt{49}}{2 \cdot 15}

x_{1} = \frac{13 - \sqrt{49}}{2 \cdot 15}

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

\sqrt{49}

7

;

7^{2}

49

ⅰ

2 \cdot 15

2 \cdot (10 + 5)

2 \cdot 10 + 2 \cdot 5

20 + 10

30

x_{1} = \frac{13 - 7}{30}

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

13 - 7

6

x_{1} = \frac{6}{30}

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

6

, their greatest common divisor:

\frac{6}{30}

\frac{1 \cdot 6}{5 \cdot 6}

\frac{1}{5}

x_{1} = \frac{1}{5}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-13) + \sqrt{49}}{2 \cdot 15}

x_{2} = \frac{13 + \sqrt{49}}{2 \cdot 15}

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

\sqrt{49}

7

;

7^{2}

49

ⅰ

2 \cdot 15

2 \cdot (10 + 5)

2 \cdot 10 + 2 \cdot 5

20 + 10

30

x_{2} = \frac{13 + 7}{30}

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

13 + 7

20

x_{2} = \frac{20}{30}

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

10

, their greatest common divisor:

\frac{20}{30}

\frac{2 \cdot 10}{3 \cdot 10}

\frac{2}{3}

x_{2} = \frac{2}{3}

A trinomial with roots

x_{1}

and

x_{2}

a \cdot (x - x_{1}) \cdot (x - x_{2})

; bringing the coefficient into the brackets:

15 \cdot (x - \frac{1}{5}) \cdot (x - \frac{2}{3})

(3 x - 2) \cdot (5 x - 1)

▶Factors identified:

(3 x - 2) \cdot (5 x - 1)