What is the completely factored form of d⁴ - 8d² + 16?

Solution:

Given d⁴ - 8d² + 16

Let a = d²

Then the polynomial is rewritten using this substitution as a² - 8a + 16.

Let us factor this quadratic polynomial by splitting the middle terms.

a² - 4a - 4a + 16

a(a-4)-4(a-4)

(a-4)(a-4)

a = 4,4

d² = 4 ⇒ d = ± 2, ± 2

d = ±2, ±2 are the roots of the equation d⁴ - 8d² + 16=0

d⁴ - 8d² + 16 = (d - 2)(d - 2)(d + 2)(d + 2).

What is the completely factored form of d⁴ - 8d² + 16?

Summary:

The completely factored form of d⁴ - 8d² + 16 is (d - 2)(d - 2)(d + 2)(d + 2).