factor 20y^2+y-1 — step-by-step solution

Let's factorize in

ℝ

20 y^{2} + y - 1

Given the quadratic form

a y^{2} + b y + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 20

b = 1

and

c = -1

Δ = 1 - 4 \cdot 20 \cdot (-1)

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 20

80

■Multiplying by -1 gives the opposite:

- 80 \cdot (-1)

80

Δ = 1 + 80

ⓘ Start from

1

and move forward

80

on the number line:

1 + 80

81

Δ = 81

Δ > 0

, the equation has two roots:

The quadratic formula gives

y_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 1 - \sqrt{81}}{2 \cdot 20}

y_{1} = \frac{- 1 - \sqrt{81}}{2 \cdot 20}

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

\sqrt{81}

9

;

9^{2}

81

ⅰ

2 \cdot 20

40

y_{1} = \frac{- 1 - 9}{40}

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

- 1 - 9

-10

y_{1} = - \frac{10}{40}

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

10

, their greatest common divisor:

- \frac{10}{40}

- \frac{1 \cdot 10}{4 \cdot 10}

- \frac{1}{4}

y_{1} = - \frac{1}{4}

The quadratic formula gives

y_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 1 + \sqrt{81}}{2 \cdot 20}

y_{2} = \frac{- 1 + \sqrt{81}}{2 \cdot 20}

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

\sqrt{81}

9

;

9^{2}

81

ⅰ

2 \cdot 20

40

y_{2} = \frac{- 1 + 9}{40}

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

- 1 + 9

8

y_{2} = \frac{8}{40}

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

8

, their greatest common divisor:

\frac{8}{40}

\frac{1 \cdot 8}{5 \cdot 8}

\frac{1}{5}

y_{2} = \frac{1}{5}

A trinomial with roots

y_{1}

and

y_{2}

a \cdot (y - y_{1}) \cdot (y - y_{2})

; bringing the coefficient into the brackets:

20 \cdot (y + \frac{1}{4}) \cdot (y - \frac{1}{5})

(4 y + 1) \cdot (5 y - 1)

▶Factors found:

(4 y + 1) \cdot (5 y - 1)