Question

difficulty: ★★★★
factorise completely $x^2 + 2xy - 4y - 4$
a $(x - 2)(y + 2)$
b $(x + 2)(y - 2)$
c $(x + 2y + 2)(x - 2)$
d $(x + 2y - 2)(x + 2)$

Explanation:

Step1: Group terms

Group the first two terms and the last two terms: \( (x^2 + 2xy) + (-4y - 4) \)

Step2: Factor each group

Factor \( x \) from the first group and \( -4 \) from the second group: \( x(x + 2y) - 4(y + 1) \) Wait, that's not right. Let's try another grouping. Wait, maybe group as \( x^2 + 2xy - 4y - 4 = x^2 + 2xy - 4(y + 1) \). No, better to rearrange or use another method. Wait, let's try factoring by grouping correctly. Let's take \( x^2 + 2xy - 4y - 4 \). Group as \( (x^2 + 2xy) + (-4y - 4) = x(x + 2y) - 4(y + 1) \). Hmm, not helpful. Wait, maybe another approach. Let's expand the options to see which one matches.

Let's expand option C: \( (x + 2y + 2)(x - 2) = x(x - 2) + 2y(x - 2) + 2(x - 2) = x^2 - 2x + 2xy - 4y + 2x - 4 = x^2 + 2xy - 4y - 4 \). Wait, that's the original expression! Wait, no, let's check again. Wait, \( (x + 2y + 2)(x - 2) = x^2 - 2x + 2xy - 4y + 2x - 4 \). The -2x and +2x cancel, so we get \( x^2 + 2xy - 4y - 4 \), which is the original. Wait, but let's check option D: \( (x + 2y - 2)(x + 2) = x(x + 2) + 2y(x + 2) - 2(x + 2) = x^2 + 2x + 2xy + 4y - 2x - 4 = x^2 + 2xy + 4y - 4 \). No, that's not. Option C: Wait, when we expand \( (x + 2y + 2)(x - 2) \), we have \( x^2 - 2x + 2xy - 4y + 2x - 4 \). The -2x and +2x cancel, so \( x^2 + 2xy - 4y - 4 \), which matches. Wait, but let's do factoring by grouping correctly. Let's take \( x^2 + 2xy - 4y - 4 \). Let's group as \( (x^2 - 4) + (2xy - 4y) \). Then \( (x - 2)(x + 2) + 2y(x - 2) \). Now factor out \( (x - 2) \): \( (x - 2)(x + 2 + 2y) \), which is \( (x - 2)(x + 2y + 2) \), which is option C.

Step3: Verify

Expanding \( (x + 2y + 2)(x - 2) \) gives \( x^2 - 2x + 2xy - 4y + 2x - 4 = x^2 + 2xy - 4y - 4 \), which matches the original expression.

Answer:

C. \( (x + 2y + 2)(x - 2) \)