factor 2j^2+17j+8 — step-by-step solution

Let's factor in

ℝ

2 j^{2} + 17 j + 8

Given the quadratic form

a j^{2} + b j + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 2

b = 17

and

c = 8

Δ = 17^{2} - 4 \cdot 2 \cdot 8

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

17^{2}

17 \cdot 17

289

■Multiplications are done before additions and subtractions:

4 \cdot 2

8

Δ = 289 - 8 \cdot 8

ⓘ Multiplications are done before additions and subtractions:

8 \cdot 8

64

Δ = 289 - 64

ⓘ Start from

289

and move back

64

on the number line:

289 - 64

225

Δ = 225

Δ > 0

, the equation has two roots:

The quadratic formula gives

j_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 17 - \sqrt{225}}{2 \cdot 2}

j_{1} = \frac{- 17 - \sqrt{225}}{2 \cdot 2}

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

\sqrt{225}

15

;

15^{2}

225

ⅰ

2 \cdot 2

4

j_{1} = \frac{- 17 - 15}{4}

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

- 17 - 15

-32

j_{1} = - \frac{32}{4}

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

4

, their greatest common divisor:

- \frac{32}{4}

-8

j_{1} = -8

The quadratic formula gives

j_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 17 + \sqrt{225}}{2 \cdot 2}

j_{2} = \frac{- 17 + \sqrt{225}}{2 \cdot 2}

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

\sqrt{225}

15

;

15^{2}

225

ⅰ

2 \cdot 2

4

j_{2} = \frac{- 17 + 15}{4}

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

- 17 + 15

-2

j_{2} = - \frac{2}{4}

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

2

, their greatest common divisor:

- \frac{2}{4}

- \frac{1 \cdot 2}{2 \cdot 2}

- \frac{1}{2}

j_{2} = - \frac{1}{2}

A trinomial with roots

j_{1}

and

j_{2}

a \cdot (j - j_{1}) \cdot (j - j_{2})

; bringing the coefficient into the brackets:

2 \cdot (j + 8) \cdot (j + \frac{1}{2})

(j + 8) \cdot (2 j + 1)

▶Factors found:

(j + 8) \cdot (2 j + 1)