Question

difficulty: ★★★
factorise completely ((x^2 + 4x)^2 + 4(x^2 + 4x) + 4).
a ((x + 2)^4)
b ((x + 4)^4)
c ((x^2 + 2x + 4)^2)
d ((x^2 + 4x + 2)^2)

Explanation:

Step1: Let \( y = x^{2}+4x \)

Substitute \( y \) into the expression, we get \( y^{2}+4y + 4 \)

Step2: Factor the quadratic in \( y \)

Notice that \( y^{2}+4y + 4=(y + 2)^{2}\) (using the perfect square formula \( a^{2}+2ab + b^{2}=(a + b)^{2} \), here \( a = y \), \( b = 2 \))

Step3: Substitute back \( y=x^{2}+4x \)

We have \( (x^{2}+4x + 2)^{2} \)

Answer:

D. \( (x^{2}+4x + 2)^{2} \)