factor x^6-16x^2 — step-by-step solution

1) Let's factorize in

ℝ

x^{6} - 16 x^{2}

For

x^{6} - 16 x^{2}

, we recognize a difference of two squares:

{(x^{3})}^{2} - {(4 x)}^{2}

. Apply

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

, with

a = x^{3}

and

b = 4 x

You can factorize the expression:

x^{6} - 16 x^{2}

into:

(x^{3} - 4 x) \cdot (x^{3} + 4 x)

2) Let's factorize in

ℝ

(x^{3} - 4 x) \cdot (x^{3} + 4 x)

⇥2.1) Let's factor in

ℝ

(x^{3} - 4 x)

⇥You can factorize this expression into:

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

⇥Let's factorize in

ℝ

(x^{2} - 4)

⇥For

x^{2} - 4

, we recognize a difference of two squares:

x^{2} - 2^{2}

. Apply

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

, with

a = x

and

b = 2

⇥Factors found:

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

▷The factor

(x^{3} - 4 x)

factors as:

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

⇥2.2) Let's factorize in

ℝ

(x^{3} + 4 x)

⇥You can factorize this expression into:

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

▷The factor

(x^{3} + 4 x)

factors as:

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

3) Let's replace each factor by its factorization:

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

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Factor out the common factor

x^{2}

■A square being never negative,

x^{2} + 4

is never zero and cannot be factored in ℝ

ⅰ

(x^{3} - 4 x) \cdot (x^{3} + 4 x)

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

■

x^{6} - 16 x^{2}

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

▶Factored expression:

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