factor 4p^2-4p-3 — step-by-step solution

Let's factorize in

ℝ

4 p^{2} - 4 p - 3

Given the quadratic form

a p^{2} + b p + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 4

b = -4

and

c = -3

Δ = {(-4)}^{2} - 4 \cdot 4 \cdot (-3)

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

{(-4)}^{2}

(-4) \cdot (-4)

16

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

■Two minus signs cancel out, the product of two negative numbers is positive:

- 16 \cdot (-3)

16 \cdot 3

Δ = 16 + 16 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

16 \cdot 3

(10 + 6) \cdot 3

10 \cdot 3 + 6 \cdot 3

30 + 18

48

Δ = 16 + 48

ⓘ Start from

16

and move forward

48

on the number line:

16 + 48

64

Δ = 64

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

p_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-4) - \sqrt{64}}{2 \cdot 4}

p_{1} = \frac{4 - \sqrt{64}}{2 \cdot 4}

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

\sqrt{64}

8

;

8^{2}

64

ⅰ

2 \cdot 4

8

p_{1} = \frac{4 - 8}{8}

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

4 - 8

-4

p_{1} = - \frac{4}{8}

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

4

, their greatest common divisor:

- \frac{4}{8}

- \frac{1 \cdot 4}{2 \cdot 4}

- \frac{1}{2}

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

The quadratic formula gives

p_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-4) + \sqrt{64}}{2 \cdot 4}

p_{2} = \frac{4 + \sqrt{64}}{2 \cdot 4}

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

\sqrt{64}

8

;

8^{2}

64

ⅰ

2 \cdot 4

8

p_{2} = \frac{4 + 8}{8}

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

4 + 8

12

p_{2} = \frac{12}{8}

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

4

, their greatest common divisor:

\frac{12}{8}

\frac{3 \cdot 4}{2 \cdot 4}

\frac{3}{2}

p_{2} = \frac{3}{2}

A trinomial with roots

p_{1}

and

p_{2}

a \cdot (p - p_{1}) \cdot (p - p_{2})

; bringing the coefficient into the brackets:

4 \cdot (p + \frac{1}{2}) \cdot (p - \frac{3}{2})

(2 p + 1) \cdot (2 p - 3)

▶Factors found:

(2 p + 1) \cdot (2 p - 3)