factor 1-16x^4 — step-by-step solution

1) Let's factorize in

ℝ

1 - 16 x^{4}

For

1 - 16 x^{4}

, we recognize a difference of two squares:

1^{2} - {(4 x^{2})}^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 1

and

b = 4 x^{2}

You can factorize the expression:

1 - 16 x^{4}

into:

(1 - 4 x^{2}) \cdot (1 + 4 x^{2})

2) Let's factor in

ℝ

(1 - 4 x^{2}) \cdot (1 + 4 x^{2})

⇥2.1) Let's factor in

ℝ

(1 - 4 x^{2})

⇥For

1 - 4 x^{2}

, we recognize a difference of two squares:

1^{2} - {(2 x)}^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 1

and

b = 2 x

⇥You can factorize the expression:

1 - 4 x^{2}

into:

(1 - 2 x) \cdot (1 + 2 x)

▷The factor

(1 - 4 x^{2})

factors as:

(1 - 2 x) \cdot (2 x + 1)

⇥2.2) Let's factorize in

ℝ

(1 + 4 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 = 4

b = 0

and

c = 1

⇥

Δ = 0^{2} - 4 \cdot 4 \cdot 1

⇥ⓘ Adding 0 changes nothing:

0 - 4 \cdot 4

- 4 \cdot 4

⇥

Δ = - 4 \cdot 4

⇥ⓘ A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

- 4 \cdot 4

-16

⇥

Δ = -16

⇥As

Δ < 0

, the equation has no real root.

⇥Factor identified:

4 x^{2} + 1

▶Factored expression:

(1 - 2 x) \cdot (2 x + 1) \cdot (4 x^{2} + 1)